Precalculus
An Investigation of Functions
Edition 1.4
David Lippman Melonie Rasmussen This book is also available to read free online at http://www.opentextbookstore.com/precalc/ If you want a printed copy, buying from the bookstore is cheaper than printing yourself.
ii Copyright © 2014 David Lippman and Melonie Rasmussen This text is licensed under a Creative Commons Attribution-Share Alike 3.0 United States License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. You are free: to Share — to copy, distribute, display, and perform the work to Remix — to make derivative works Under the following conditions: Attribution. You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). Share Alike. If you alter, transform, or build upon this work, you may distribute the resulting work only under the same, similar or a compatible license. With the understanding that: Waiver. Any of the above conditions can be waived if you get permission from the copyright holder. Other Rights. In no way are any of the following rights affected by the license: • Your fair dealing or fair use rights; • Apart from the remix rights granted under this license, the author's moral rights; • Rights other persons may have either in the work itself or in how the work is used, such as publicity or privacy rights. • Notice — For any reuse or distribution, you must make clear to others the license terms of this work. The best way to do this is with a link to this web page: http://creativecommons.org/licenses/by-sa/3.0/us/
In addition to these rights, we give explicit permission to remix small portions of this book (less than 10% cumulative) into works that are CC-BY, CC-BY-SA-NC, or GFDL licensed. Selected exercises were remixed from Precalculus by D.H. Collingwood and K.D. Prince, originally licensed under the GNU Free Document License, with permission from the authors. Cover Photo by David Lippman, of artwork by John Rogers Lituus, 2010 Dichromatic glass and aluminum Washington State Arts Commission in partnership with Pierce College
This is the fifth official version of Edition 1. It contains typo corrections and language clarification, but is page number and problem set number equivalent to the original Edition 1.
i
About the Authors David Lippman received his master’s degree in mathematics from Western Washington University and has been teaching at Pierce College since Fall 2000. Melonie Rasmussen also received her master’s degree in mathematics from Western Washington University and has been teaching at Pierce College since Fall 2002. Prior to this Melonie taught for the Puyallup School district for 6 years after receiving her teaching credentials from Pacific Lutheran University. We have both been long time advocates of open learning, open materials, and basically any idea that will reduce the cost of education for students. It started by supporting the college’s calculator rental program, and running a book loan scholarship program. Eventually the frustration with the escalating costs of commercial text books and the online homework systems that charged for access led them to take action. First, David developed IMathAS, open source online math homework software that runs WAMAP.org and MyOpenMath.com. Through this platform, we became integral parts of a vibrant sharing and learning community of teachers from around Washington State that support and contribute to WAMAP. Our pioneering efforts, supported by dozens of other dedicated faculty and financial support from the Transition Math Project, have led to a system used by thousands of students every quarter, saving hundreds of thousands of dollars over comparable commercial offerings. David continued further and wrote his first open textbook, Math in Society, a math for liberal arts majors book, after being frustrated by students having to pay $100+ for a textbook for a terminal course. Together, frustrated by both cost and the style of commercial texts, we began writing PreCalculus: An Investigation of Functions in 2010.
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Acknowledgements We would like to thank the following for their generous support and feedback. •
The community of WAMAP users and developers for creating a majority of the homework content used in our online homework sets.
•
Pierce College students in our Fall 2010 - Summer 2011 Math 141 and Math 142 classes for helping correct typos, identifying videos related to the homework, and being our willing test subjects.
•
The Open Course Library Project for providing the support needed to produce a full course package for these courses.
•
Mike Kenyon, Chris Willett, Tophe Anderson, and Vauhn Foster-Grahler for reviewing the course and giving feedback and suggestions.
•
Our Pierce College colleagues for providing their suggestions.
•
Tophe Anderson, James Gray, and Lawrence Morales for their feedback and suggestions in content and examples.
•
Jeff Eldridge for extensive proofreading and suggestions for clarification.
•
James Sousa for developing videos associated with the online homework.
•
Kevin Dimond for his work on indexing the book and creating PowerPoint slides.
•
Faculty at Green River Community College and the Maricopa College District for their feedback and suggestions.
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Preface Over the years, when reviewing books we found that many had been mainstreamed by the publishers in an effort to appeal to everyone, leaving them with very little character. There were only a handful of books that had the conceptual and application driven focus we liked, and most of those were lacking in other aspects we cared about, like providing students sufficient examples and practice of basic skills. The largest frustration, however, was the never ending escalation of cost and being forced into new editions every three years. We began researching open textbooks, however the ability for those books to be adapted, remixed, or printed were often limited by the types of licenses, or didn’t approach the material the way we wanted. This book is available online for free, in both Word and PDF format. You are free to change the wording, add materials and sections or take them away. We welcome feedback, comments and suggestions for future development at
[email protected]. Additionally, if you add a section, chapter or problems, we would love to hear from you and possibly add your materials so everyone can benefit. In writing this book, our focus was on the story of functions. We begin with function notation, a basic toolkit of functions, and the basic operation with functions: composition and transformation. Building from these basic functions, as each new family of functions is introduced we explore the important features of the function: its graph, domain and range, intercepts, and asymptotes. The exploration then moves to evaluating and solving equations involving the function, finding inverses, and culminates with modeling using the function. The "rule of four" is integrated throughout - looking at the functions verbally, graphically, numerically, as well as algebraically. We feel that using the “rule of four” gives students the tools they need to approach new problems from various angles. Often the “story problems of life” do not always come packaged in a neat equation. Being able to think critically, see the parts and build a table or graph a trend, helps us change the words into meaningful and measurable functions that model the world around us. There is nothing we hate more than a chapter on exponential equations that begins "Exponential functions are functions that have the form f(x)=ax." As each family of functions is introduced, we motivate the topic by looking at how the function arises from life scenarios or from modeling. Also, we feel it is important that precalculus be the bridge in level of thinking between algebra and calculus. In algebra, it is common to see numerous examples with very similar homework exercises, encouraging the student to mimic the examples. Precalculus provides a link that takes students from the basic plug & chug of formulaic calculations towards building an understanding that equations and formulas have deeper meaning and purpose. While you will find examples and similar exercises for the basic skills in this book, you will also find examples of multistep problem solving along with exercises in multistep problem solving. Often times these exercises will not exactly mimic the exercises, forcing the students to employ their critical thinking skills and apply the skills they've learned to new situations. By
iv developing students’ critical thinking and problem solving skills this course prepares students for the rigors of Calculus. While we followed a fairly standard ordering of material in the first half of the book, we took some liberties in the trig portion of the book. It is our opinion that there is no need to separate unit circle trig from triangle trig, and instead integrated them in the first chapter. Identities are introduced in the first chapter, and revisited throughout. Likewise, solving is introduced in the second chapter and revisited more extensively in the third chapter. As with the first part of the book, an emphasis is placed on motivating the concepts and on modeling and interpretation.
Supplements During Spring 2010, the Washington Open Course Library (OCL) project was announced with the goal of creating open courseware for the 81 highest enrolled community college courses with a price cap on course materials of $30. We were chosen to work on precalculus for this project, and that helped drive us to complete our book, and allowed us to create supplemental materials. A course package is available that contains the following features: • • • • • • • • • • • •
Suggested syllabus Day by day course guide Instructor guide with lecture outlines and examples Additional online resources, with links to other textbooks, videos, and other resources Discussion forums Diagnostic review Online homework for each section (algorithmically generated, free response) A list of videos related to the online homework Printable class worksheets, activities, and handouts Chapter review problems Sample quizzes Sample chapter exams
The course shell was built for the IMathAS online homework platform, and is available for Washington State faculty at www.wamap.org and mirrored for others at www.myopenmath.com. The course shell was designed to follow Quality Matters (QM) guidelines, but has not yet been formally reviewed.
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How To Be Successful In This Course
This is not a high school math course, although for some of you the content may seem familiar. There are key differences to what you will learn here, how quickly you will be required to learn it and how much work will be required of you. You will no longer be shown a technique and be asked to mimic it repetitively as the only way to prove learning. Not only will you be required to master the technique, but you will also be required to extend that knowledge to new situations and build bridges between the material at hand and the next topic, making the course highly cumulative. As a rule of thumb, for each hour you spend in class, you should expect this course will require an average of 2 hours of out-of-class focused study. This means that some of you with a stronger background in mathematics may take less, but if you have a weaker background or any math anxiety it will take you more. Notice how this is the equivalent of having a part time job, and if you are taking a fulltime load of courses as many college students do, this equates to more than a full time job. If you must work, raise a family and take a full load of courses all at the same time, we recommend that you get a head start & get organized as soon as possible. We also recommend that you spread out your learning into daily chunks and avoid trying to cram or learn material quickly before an exam. To be prepared, read through the material before it is covered in class and note or highlight the material that is new or confusing. The instructor’s lecture and activities should not be the first exposure to the material. As you read, test your understanding with the Try it Now problems in the book. If you can’t figure one out, try again after class, and ask for help if you still can’t get it. As soon as possible after the class session recap the day’s lecture or activities into a meaningful format to provide a third exposure to the material. You could summarize your notes into a list of key points, or reread your notes and try to work examples done in class without referring back to your notes. Next, begin any assigned homework. The next day, if the instructor provides the opportunity to clarify topics or ask questions, do not be afraid to ask. If you are afraid to ask, then you are not getting your money’s worth! If the instructor does not provide this opportunity, be prepared to go to a tutoring center or build a peer study group. Put in quality effort and time and you can get quality results. Lastly, if you feel like you do not understand a topic. Don’t wait, ASK FOR HELP! ASK: Ask a teacher or tutor, Search for ancillaries, Keep a detailed list of questions FOR: Find additional resources, Organize the material, Research other learning options HELP: Have a support network, Examine your weaknesses, List specific examples & Practice
Best of luck learning! We hope you like the course & love the price. David & Melonie
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Table of Contents About the Authors ............................................................................................................ i Acknowledgements ......................................................................................................... ii Preface ........................................................................................................................... iii Supplements ................................................................................................................... iv How To Be Successful In This Course ........................................................................... v Table of Contents ........................................................................................................... vi Chapter 1: Functions ........................................................................................................ 1 Section 1.1 Functions and Function Notation ................................................................. 1 Section 1.2 Domain and Range..................................................................................... 21 Section 1.3 Rates of Change and Behavior of Graphs .................................................. 34 Section 1.4 Composition of Functions .......................................................................... 49 Section 1.5 Transformation of Functions ..................................................................... 61 Section 1.6 Inverse Functions ....................................................................................... 90 Chapter 2: Linear Functions.......................................................................................... 99 Section 2.1 Linear Functions ........................................................................................ 99 Section 2.2 Graphs of Linear Functions ..................................................................... 111 Section 2.3 Modeling with Linear Functions.............................................................. 126 Section 2.4 Fitting Linear Models to Data .................................................................. 138 Section 2.5 Absolute Value Functions ........................................................................ 146 Chapter 3: Polynomial and Rational Functions......................................................... 155 Section 3.1 Power Functions & Polynomial Functions .............................................. 155 Section 3.2 Quadratic Functions ................................................................................. 163 Section 3.3 Graphs of Polynomial Functions ............................................................. 176 Section 3.4 Rational Functions ................................................................................... 188 Section 3.5 Inverses and Radical Functions ............................................................... 206 Chapter 4: Exponential and Logarithmic Functions................................................. 215 Section 4.1 Exponential Functions ............................................................................. 215 Section 4.2 Graphs of Exponential Functions ............................................................ 232 Section 4.3 Logarithmic Functions ............................................................................. 242 Section 4.4 Logarithmic Properties ............................................................................ 253 Section 4.5 Graphs of Logarithmic Functions ............................................................ 262 Section 4.6 Exponential and Logarithmic Models...................................................... 270 Section 4.7 Fitting Exponentials to Data .................................................................... 289
vii Chapter 5: Trigonometric Functions of Angles ......................................................... 297 Section 5.1 Circles ...................................................................................................... 297 Section 5.2 Angles ...................................................................................................... 307 Section 5.3 Points on Circles using Sine and Cosine ................................................. 321 Section 5.4 The Other Trigonometric Functions ........................................................ 333 Section 5.5 Right Triangle Trigonometry ................................................................... 343 Chapter 6: Periodic Functions ..................................................................................... 353 Section 6.1 Sinusoidal Graphs .................................................................................... 353 Section 6.2 Graphs of the Other Trig Functions ......................................................... 369 Section 6.3 Inverse Trig Functions ............................................................................. 379 Section 6.4 Solving Trig Equations ............................................................................ 387 Section 6.5 Modeling with Trigonometric Equations ................................................. 397 Chapter 7: Trigonometric Equations and Identities ................................................. 409 Section 7.1 Solving Trigonometric Equations with Identities .................................... 409 Section 7.2 Addition and Subtraction Identities ......................................................... 417 Section 7.3 Double Angle Identities ........................................................................... 431 Section 7.4 Modeling Changing Amplitude and Midline ........................................... 442 Chapter 8: Further Applications of Trigonometry.................................................... 451 Section 8.1 Non-right Triangles: Law of Sines and Cosines ...................................... 451 Section 8.2 Polar Coordinates ..................................................................................... 467 Section 8.3 Polar Form of Complex Numbers ............................................................ 480 Section 8.4 Vectors ..................................................................................................... 491 Section 8.5 Parametric Equations ............................................................................... 504 Solutions to Selected Exercises .................................................................................... 519 Chapter 1 ..................................................................................................................... 519 Chapter 2 ..................................................................................................................... 526 Chapter 3 ..................................................................................................................... 530 Chapter 4 ..................................................................................................................... 534 Chapter 5 ..................................................................................................................... 539 Chapter 6 ..................................................................................................................... 542 Chapter 7 ..................................................................................................................... 546 Chapter 8 ..................................................................................................................... 549 Index ............................................................................................................................... 555
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Section 1.1 Functions and Function Notation
1
Chapter 1: Functions
Section 1.1 Functions and Function Notation ................................................................. 1 Section 1.2 Domain and Range ..................................................................................... 21 Section 1.3 Rates of Change and Behavior of Graphs .................................................. 34 Section 1.4 Composition of Functions .......................................................................... 49 Section 1.5 Transformation of Functions...................................................................... 61 Section 1.6 Inverse Functions ....................................................................................... 90
Section 1.1 Functions and Function Notation What is a Function? The natural world is full of relationships between quantities that change. When we see these relationships, it is natural for us to ask “If I know one quantity, can I then determine the other?” This establishes the idea of an input quantity, or independent variable, and a corresponding output quantity, or dependent variable. From this we get the notion of a functional relationship in which the output can be determined from the input. For some quantities, like height and age, there are certainly relationships between these quantities. Given a specific person and any age, it is easy enough to determine their height, but if we tried to reverse that relationship and determine age from a given height, that would be problematic, since most people maintain the same height for many years. Function Function: A rule for a relationship between an input, or independent, quantity and an output, or dependent, quantity in which each input value uniquely determines one output value. We say “the output is a function of the input.” Example 1 In the height and age example above, is height a function of age? Is age a function of height? In the height and age example above, it would be correct to say that height is a function of age, since each age uniquely determines a height. For example, on my 18th birthday, I had exactly one height of 69 inches. However, age is not a function of height, since one height input might correspond with more than one output age. For example, for an input height of 70 inches, there is more than one output of age since I was 70 inches at the age of 20 and 21.
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
2 Chapter 1 Example 2 At a coffee shop, the menu consists of items and their prices. Is price a function of the item? Is the item a function of the price? We could say that price is a function of the item, since each input of an item has one output of a price corresponding to it. We could not say that item is a function of price, since two items might have the same price. Example 3 In many classes the overall percentage you earn in the course corresponds to a decimal grade point. Is decimal grade a function of percentage? Is percentage a function of decimal grade? For any percentage earned, there would be a decimal grade associated, so we could say that the decimal grade is a function of percentage. That is, if you input the percentage, your output would be a decimal grade. Percentage may or may not be a function of decimal grade, depending upon the teacher’s grading scheme. With some grading systems, there are a range of percentages that correspond to the same decimal grade. One-to-One Function Sometimes in a relationship each input corresponds to exactly one output, and every output corresponds to exactly one input. We call this kind of relationship a one-to-one function. From Example 3, if each unique percentage corresponds to one unique decimal grade point and each unique decimal grade point corresponds to one unique percentage then it is a one-to-one function. Try it Now Let’s consider bank account information. 1. Is your balance a function of your bank account number?
(if you input a bank account number does it make sense that the output is your balance?)
2. Is your bank account number a function of your balance?
(if you input a balance does it make sense that the output is your bank account number?)
Function Notation To simplify writing out expressions and equations involving functions, a simplified notation is often used. We also use descriptive variables to help us remember the meaning of the quantities in the problem.
Section 1.1 Functions and Function Notation
3
Rather than write “height is a function of age”, we could use the descriptive variable h to represent height and we could use the descriptive variable a to represent age. “height is a function of age” “h is f of a” h = f(a) h(a)
if we name the function f we write or more simply we could instead name the function h and write which is read “h of a”
Remember we can use any variable to name the function; the notation h(a) shows us that h depends on a. The value “a” must be put into the function “h” to get a result. Be careful - the parentheses indicate that age is input into the function (Note: do not confuse these parentheses with multiplication!). Function Notation The notation output = f(input) defines a function named f. This would be read “output is f of input” Example 4 Introduce function notation to represent a function that takes as input the name of a month, and gives as output the number of days in that month. The number of days in a month is a function of the name of the month, so if we name the function f, we could write “days = f(month)” or d = f(m). If we simply name the function d, we could write d(m) For example, d(March) = 31, since March has 31 days. The notation d(m) reminds us that the number of days, d (the output) is dependent on the name of the month, m (the input) Example 5 A function N = f(y) gives the number of police officers, N, in a town in year y. What does f(2005) = 300 tell us? When we read f(2005) = 300, we see the input quantity is 2005, which is a value for the input quantity of the function, the year (y). The output value is 300, the number of police officers (N), a value for the output quantity. Remember N=f(y). So this tells us that in the year 2005 there were 300 police officers in the town. Tables as Functions Functions can be represented in many ways: Words (as we did in the last few examples), tables of values, graphs, or formulas. Represented as a table, we are presented with a list of input and output values.
4 Chapter 1 In some cases, these values represent everything we know about the relationship, while in other cases the table is simply providing us a few select values from a more complete relationship. Table 1: This table represents the input, number of the month (January = 1, February = 2, and so on) while the output is the number of days in that month. This represents everything we know about the months & days for a given year (that is not a leap year) (input) Month number, m (output) Days in month, D
1
2
3
4
5
6
7
8
9
10
11
12
31
28
31
30
31
30
31
31
30
31
30
31
Table 2: The table below defines a function Q = g(n). Remember this notation tells us g is the name of the function that takes the input n and gives the output Q. n Q
1 8
2 6
3 7
4 6
5 8
Table 3: This table represents the age of children in years and their corresponding heights. This represents just some of the data available for height and ages of children. (input) a, age in years (output) h, height inches
5
5
6
7
8
9
10
40
42
44
47
50
52
54
Example 6 Which of these tables define a function (if any)? Are any of them one-to-one? Input 2 5 8
Output 1 3 6
Input -3 0 4
Output 5 1 5
Input 1 5 5
Output 0 2 4
The first and second tables define functions. In both, each input corresponds to exactly one output. The third table does not define a function since the input value of 5 corresponds with two different output values. Only the first table is one-to-one; it is both a function, and each output corresponds to exactly one input. Although table 2 is a function, because each input corresponds to exactly one output, each output does not correspond to exactly one input so this function is not one-to-one. Table 3 is not even a function and so we don’t even need to consider if it is a one-to-one function.
Section 1.1 Functions and Function Notation Try it Now 3. If each percentage earned translated to one letter grade, would this be a function? Is it one-to-one? Solving and Evaluating Functions: When we work with functions, there are two typical things we do: evaluate and solve. Evaluating a function is what we do when we know an input, and use the function to determine the corresponding output. Evaluating will always produce one result, since each input of a function corresponds to exactly one output. Solving equations involving a function is what we do when we know an output, and use the function to determine the inputs that would produce that output. Solving a function could produce more than one solution, since different inputs can produce the same output. Example 7 Using the table shown, where Q=g(n) a) Evaluate g(3)
n Q
1 8
2 6
3 7
4 6
5 8
Evaluating g(3) (read: “g of 3”) means that we need to determine the output value, Q, of the function g given the input value of n=3. Looking at the table, we see the output corresponding to n=3 is Q=7, allowing us to conclude g(3) = 7. b) Solve g(n) = 6 Solving g(n) = 6 means we need to determine what input values, n, produce an output value of 6. Looking at the table we see there are two solutions: n = 2 and n = 4. When we input 2 into the function g, our output is Q = 6 When we input 4 into the function g, our output is also Q = 6 Try it Now 4. Using the function in Example 7, evaluate g(4) Graphs as Functions Oftentimes a graph of a relationship can be used to define a function. By convention, graphs are typically created with the input quantity along the horizontal axis and the output quantity along the vertical.
5
6 Chapter 1 The most common graph has y on the vertical axis and x on the horizontal axis, and we say y is a function of x, or y = f(x) when the function is named f. y
x
Example 8 Which of these graphs defines a function y=f(x)? Which of these graphs defines a oneto-one function?
Looking at the three graphs above, the first two define a function y=f(x), since for each input value along the horizontal axis there is exactly one output value corresponding, determined by the y-value of the graph. The 3rd graph does not define a function y=f(x) since some input values, such as x=2, correspond with more than one output value. Graph 1 is not a one-to-one function. For example, the output value 3 has two corresponding input values, -2 and 2.3 Graph 2 is a one-to-one function; each input corresponds to exactly one output, and every output corresponds to exactly one input. Graph 3 is not even a function so there is no reason to even check to see if it is a one-toone function. Vertical Line Test The vertical line test is a handy way to think about whether a graph defines the vertical output as a function of the horizontal input. Imagine drawing vertical lines through the graph. If any vertical line would cross the graph more than once, then the graph does not define only one vertical output for each horizontal input.
Section 1.1 Functions and Function Notation
7
Horizontal Line Test Once you have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line crosses the graph more than once, then the graph does not define a one-to-one function. Evaluating a function using a graph requires taking the given input and using the graph to look up the corresponding output. Solving a function equation using a graph requires taking the given output and looking on the graph to determine the corresponding input. Example 9 Given the graph below, a) Evaluate f(2) b) Solve f(x) = 4
a) To evaluate f(2), we find the input of x=2 on the horizontal axis. Moving up to the graph gives the point (2, 1), giving an output of y=1. So f(2) = 1 b) To solve f(x) = 4, we find the value 4 on the vertical axis because if f(x) = 4 then 4 is the output. Moving horizontally across the graph gives two points with the output of 4: (-1,4) and (3,4). These give the two solutions to f(x) = 4: x = -1 or x = 3 This means f(-1)=4 and f(3)=4, or when the input is -1 or 3, the output is 4. Notice that while the graph in the previous example is a function, getting two input values for the output value of 4 shows us that this function is not one-to-one. Try it Now 5. Using the graph from example 9, solve f(x)=1. Formulas as Functions When possible, it is very convenient to define relationships using formulas. If it is possible to express the output as a formula involving the input quantity, then we can define a function.
8 Chapter 1 Example 10 Express the relationship 2n + 6p = 12 as a function p = f(n) if possible. To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [something involving n]. 2n + 6p = 12 6p = 12 - 2n
subtract 2n from both sides divide both sides by 6 and simplify
12 − 2n 12 2n 1 p= = − =2 − n 6 6 6 3
Having rewritten the formula as p=, we can now express p as a function: 1 p= f (n)= 2 − n 3 It is important to note that not every relationship can be expressed as a function with a formula. Note the important feature of an equation written as a function is that the output value can be determined directly from the input by doing evaluations - no further solving is required. This allows the relationship to act as a magic box that takes an input, processes it, and returns an output. Modern technology and computers rely on these functional relationships, since the evaluation of the function can be programmed into machines, whereas solving things is much more challenging. Example 11 Express the relationship x 2 + y 2 = 1 as a function y = f(x) if possible. If we try to solve for y in this equation: y 2 = 1 − x2 y= ± 1 − x2 We end up with two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f(x) As with tables and graphs, it is common to evaluate and solve functions involving formulas. Evaluating will require replacing the input variable in the formula with the value provided and calculating. Solving will require replacing the output variable in the formula with the value provided, and solving for the input(s) that would produce that output.
Section 1.1 Functions and Function Notation
9
Example 12 Given the function k (t )= t 3 + 2 a) Evaluate k(2) b) Solve k(t) = 1 a) To evaluate k(2), we plug in the input value 2 into the formula wherever we see the input variable t, then simplify k (2)= 23 + 2 k (2)= 8 + 2 So k(2) = 10 b) To solve k(t) = 1, we set the formula for k(t) equal to 1, and solve for the input value that will produce that output k(t) = 1 substitute the original formula k (t )= t 3 + 2 subtract 2 from each side t3 + 2 = 1 3 take the cube root of each side t = −1 t = −1 When solving an equation using formulas, you can check your answer by using your solution in the original equation to see if your calculated answer is correct. We want to know if k (t ) = 1 is true when t = −1 . k (−1) =(−1)3 + 2 = −1 + 2 = 1 which was the desired result. Example 13 Given the function h( p= ) p2 + 2 p a) Evaluate h(4) b) Solve h(p) = 3 To evaluate h(4) we substitute the value 4 for the input variable p in the given function. a) h= (4) (4) 2 + 2(4) = 16 + 8 = 24 b) h(p) = 3 p2 + 2 p = 3 p2 + 2 p − 3 = 0 2 p + 2p −3 = 0 ( p + 3)( p − 1) = 0
Substitute the original function h( p= ) p2 + 2 p This is quadratic, so we can rearrange the equation to get it = 0 subtract 3 from each side this is factorable, so we factor it
10 Chapter 1 By the zero factor theorem since ( p + 3)( p − 1) = 0 , either ( p + 3) = 0 or ( p − 1) = 0 (or both of them equal 0) and so we solve both equations for p, finding p = -3 from the first equation and p = 1 from the second equation. This gives us the solution: h(p) = 3 when p = 1 or p = -3 We found two solutions in this case, which tells us this function is not one-to-one. Try it Now 6. Given the function g (m = ) a. Evaluate g(5) b. Solve g(m) = 2
m−4
Basic Toolkit Functions In this text, we will be exploring functions – the shapes of their graphs, their unique features, their equations, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of elements to build from. We call these our “toolkit of functions” – a set of basic named functions for which we know the graph, equation, and special features. For these definitions we will use x as the input variable and f(x) as the output variable. Toolkit Functions Linear Constant: Identity: Absolute Value: Power Quadratic: Cubic:
f ( x) = c , where c is a constant (number) f ( x) = x
f ( x) = x
Square root:
f ( x) = x 2 f ( x) = x 3 1 f ( x) = x 1 f ( x) = 2 x 2 f (= x) = x
Cube root:
f ( x) = 3 x
Reciprocal: Reciprocal squared:
x
Section 1.1 Functions and Function Notation 11 You will see these toolkit functions, combinations of toolkit functions, their graphs and their transformations frequently throughout this book. In order to successfully follow along later in the book, it will be very helpful if you can recognize these toolkit functions and their features quickly by name, equation, graph and basic table values. Not every important equation can be written as y = f(x). An example of this is the equation of a circle. Recall the distance formula for the distance between two points:
dist = ( x 2 − x1 ) + ( y 2 − y1 ) A circle with radius r with center at (h, k) can be described as all points (x, y) a distance 2
2
of r from the center, so using the distance formula, r =
(x − h )2 + ( y − k )2 , giving
Equation of a circle 2 2 A circle with radius r with center (h, k) has equation r 2 = ( x − h ) + ( y − k ) Graphs of the Toolkit Functions Constant Function: f ( x) = 2
Quadratic: f ( x) = x 2
Identity: f ( x) = x
Cubic: f ( x) = x 3
Absolute Value: f ( x) = x
Square root: f ( x) = x
12 Chapter 1 Cube root: f ( x) = 3 x
Reciprocal: f ( x) =
Important Topics of this Section Definition of a function Input (independent variable) Output (dependent variable) Definition of a one-to-one function Function notation Descriptive variables Functions in words, tables, graphs & formulas Vertical line test Horizontal line test Evaluating a function at a specific input value Solving a function given a specific output value Toolkit Functions Try it Now Answers 1. Yes 2. No 3. Yes it’s a function; No, it’s not one-to-one 4. Q=g(4)=6 5. x = 0 or x = 2 6. a. g(5)=1 b. m = 8
1 x
Reciprocal squared: f ( x) =
1 x2
Section 1.1 Functions and Function Notation 13
Section 1.1 Exercises 1. The amount of garbage, G, produced by a city with population p is given by G = f ( p ) . G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function f. b. Explain the meaning of the statement f ( 5 ) = 2 . 2. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D = g (a ) . a. A garden with area 5000 ft2 requires 50 cubic yards of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g (100 ) = 1 . 3. Let f (t ) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. f ( 5 ) = 30 b. f (10 ) = 40 4. Let h(t ) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: b. h ( 2 ) = 350 a. h (1) = 200 5. Select all of the following graphs which represent y as a function of x.
a
b
c
d
e
f
14 Chapter 1
6. Select all of the following graphs which represent y as a function of x.
a
b
c
d
e
f
7. Select all of the following tables which represent y as a function of x. a. x 5 10 15 b. x 5 10 15 c. x 5 10 10 y 3 8 14 y 3 8 8 y 3 8 14 8. Select all of the following tables which represent y as a function of x. a. x 2 6 13 b. x 2 6 6 c. x 2 6 13 3 10 10 3 10 14 y y y 3 10 14 9. Select all of the following tables which represent y as a function of x. a. x y b. x y c. x y d. x y 0 -2 -1 -4 0 -5 -1 -4 3 1 2 3 3 1 1 2 4 6 5 4 3 4 4 2 8 9 8 7 9 8 9 7 3 1 12 11 16 13 12 13 10. Select all of the following tables which represent y as a function of x. a. x y b. x y c. x y d. x y -4 -2 -5 -3 -1 -3 -1 -5 3 2 2 1 1 2 3 1 6 4 2 4 5 4 5 1 9 7 7 9 9 8 8 7 12 16 11 10 1 2 14 12
Section 1.1 Functions and Function Notation 15
11. Select all of the following tables which represent y as a function of x and are one-toone. a. x 3 8 12 b. x 3 8 12 c. x 3 8 8 y 4 7 7 y 4 7 13 y 4 7 13 12. Select all of the following tables which represent y as a function of x and are one-toone. a. x 2 8 8 b. x 2 8 14 c. x 2 8 14 y 5 6 13 y 5 6 6 y 5 6 13 13. Select all of the following graphs which are one-to-one functions.
a.
b.
c.
d.
e.
f.
14. Select all of the following graphs which are one-to-one functions.
a
b
c
d
e
f
16 Chapter 1 Given each function f ( x) graphed, evaluate f (1) and f (3)
15.
16.
17. Given the function g ( x) graphed here, a. Evaluate g (2) b. Solve g ( x ) = 2
18. Given the function f ( x) graphed here. a. Evaluate f ( 4 ) b. Solve f ( x) 4 =
19. Based on the table below, b. Solve f ( x) a. Evaluate f (3) =1 0 1 2 3 4 5 6 7 8 9 x f ( x ) 74 28 1 53 56 3 36 45 14 47 20. Based on the table below, a. Evaluate f (8) b. Solve f ( x) 7 = 0 1 2 3 4 5 6 7 8 9 x f ( x ) 62 8 7 38 86 73 70 39 75 34 For each of the following functions, evaluate: f ( −2 ) , f (−1) , f (0) , f (1) , and f (2) 21. f ( x )= 4 − 2 x
22. f ( x )= 8 − 3 x
x − x+ 23. f ( x ) = 8 7 3
x2 − x + 24. f ( x ) = 6 7 4
25. f ( x ) = − x3 + 2 x
26. f (= x ) 5x4 + x2
27. f ( x ) = 3+ x +3
28. f ( x ) =4 − 3 x − 2
29. f ( x ) = ( x − 2 ) ( x + 3)
30. f ( x ) = ( x + 3)( x − 1)
x −3 x +1 33. f ( x ) = 2 x
32. f ( x ) =
2
31. f ( x ) =
x−2 x+2 34. f ( x ) = 3x
2
Section 1.1 Functions and Function Notation 17 35. Suppose f ( x ) = x 2 + 8 x − 4 . Compute the following: a. f (−1) + f (1)
b. f (−1) − f (1)
36. Suppose f ( x ) = x 2 + x + 3 . Compute the following: a. f (−2) + f (4)
b. f (−2) − f (4)
37. Let f ( t = ) 3t + 5 b. Solve f ( t ) = 0
a. Evaluate f (0) 38. Let g ( p )= 6 − 2 p
b. Solve g ( p ) = 0
a. Evaluate g (0)
39. Match each function name with its equation. a. y = x i. Cube root b. y = x 3 ii. Reciprocal iii. Linear c. y = 3 x iv. Square Root 1 v. Absolute Value d. y = x vi. Quadratic e. y = x 2 vii. Reciprocal Squared viii. Cubic f. y = x g. y = x h.
y=
1 x2
40. Match each graph with its equation. i. ii. a. y = x b. y = x 3 c. y = 3 x 1 d. y = x e. y = x 2 f. y = x g. y = x h. y =
1 x2
v.
vi.
iii.
iv.
vii.
viii.
18 Chapter 1 41. Match each table with its equation. i. In a. y = x 2 -2 b. y = x -1 c. y = x 0 d. y = 1/ x 1 e. y =| x | 2 3 f. y = x 3 iv.
In -2 -1 0 1 2 3
42. Match each equation with its table a. Quadratic i. b. Absolute Value c. Square Root d. Linear e. Cubic f. Reciprocal
iv.
Out -0.5 -1 _ 1 0.5 0.33
ii.
In -2 -1 0 1 2 3
Out -2 -1 0 1 2 3
iii.
In -2 -1 0 1 2 3
Out -8 -1 0 1 8 27
Out 4 1 0 1 4 9
v.
In -2 -1 0 1 4 9
Out _ _ 0 1 2 3
vi.
In -2 -1 0 1 2 3
Out 2 1 0 1 2 3
In -2 -1 0 1 2 3
Out -0.5 -1 _ 1 0.5 0.33
ii.
In -2 -1 0 1 2 3
Out -2 -1 0 1 2 3
iii.
In -2 -1 0 1 2 3
Out -8 -1 0 1 8 27
In -2 -1 0 1 2 3
Out 4 1 0 1 4 9
v.
In -2 -1 0 1 4 9
Out _ _ 0 1 2 3
vi.
In -2 -1 0 1 2 3
Out 2 1 0 1 2 3
43. Write the equation of the circle centered at (3 , −9 ) with radius 6. 44. Write the equation of the circle centered at (9 , −8 ) with radius 11. 45. Sketch a reasonable graph for each of the following functions. [UW] a. Height of a person depending on age. b. Height of the top of your head as you jump on a pogo stick for 5 seconds. c. The amount of postage you must put on a first class letter, depending on the weight of the letter.
Section 1.1 Functions and Function Notation 19 46. Sketch a reasonable graph for each of the following functions. [UW] a. Distance of your big toe from the ground as you ride your bike for 10 seconds. b. Your height above the water level in a swimming pool after you dive off the high board. c. The percentage of dates and names you’ll remember for a history test, depending on the time you study. 47. Using the graph shown, a. Evaluate f (c) b. Solve f ( x ) = p
t r a b
L c
f(x) x
c. Suppose f ( b ) = z . Find f ( z ) d. What are the coordinates of points L and K?
K
p
48. Dave leaves his office in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function s = d(t) which keeps track of Dave’s distance s from Padelford Hall at time t. Take distance units to be “feet” and time units to be “minutes.” Assume Dave’s path to Gould Hall is long a straight line which is 2400 feet long. [UW]
a. Dave leaves Padelford Hall and walks at a constant spend until he reaches Gould Hall 10 minutes later. b. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall. c. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall.
20 Chapter 1 d. Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to figure out where he is. Dave is totally lost, so he simply heads back to his office, walking the same constant speed he had when he originally left Padelford Hall. e. Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of “distance from Padelford” vs. “time” for the both Angela and Dave. f. Suppose you want to sketch the graph of a new function s = g(t) that keeps track of Dave’s distance s from Gould Hall at time t. How would your graphs change in (a)-(e)?
Section 1.2 Domain and Range 21
Section 1.2 Domain and Range One of our main goals in mathematics is to model the real world with mathematical functions. In doing so, it is important to keep in mind the limitations of those models we create. This table shows a relationship between circumference and height of a tree as it grows. Circumference, c Height, h
1.7 24.5
2.5 31
5.5 45.2
8.2 54.6
13.7 92.1
While there is a strong relationship between the two, it would certainly be ridiculous to talk about a tree with a circumference of -3 feet, or a height of 3000 feet. When we identify limitations on the inputs and outputs of a function, we are determining the domain and range of the function. Domain and Range Domain: The set of possible input values to a function Range: The set of possible output values of a function Example 1 Using the tree table above, determine a reasonable domain and range. We could combine the data provided with our own experiences and reason to approximate the domain and range of the function h = f(c). For the domain, possible values for the input circumference c, it doesn’t make sense to have negative values, so c > 0. We could make an educated guess at a maximum reasonable value, or look up that the maximum circumference measured is about 119 feet 1. With this information we would say a reasonable domain is 0 < c ≤ 119 feet. Similarly for the range, it doesn’t make sense to have negative heights, and the maximum height of a tree could be looked up to be 379 feet, so a reasonable range is 0 < h ≤ 379 feet. Example 2 When sending a letter through the United States Postal Service, the price depends upon the weight of the letter 2, as shown in the table below. Determine the domain and range.
1 2
http://en.wikipedia.org/wiki/Tree, retrieved July 19, 2010 http://www.usps.com/prices/first-class-mail-prices.htm, retrieved July 19, 2010
22 Chapter 1 Letters Weight not Over 1 ounce 2 ounces 3 ounces 3.5 ounces
Price $0.44 $0.61 $0.78 $0.95
Suppose we notate Weight by w and Price by p, and set up a function named P, where Price, p is a function of Weight, w. p = P(w). Since acceptable weights are 3.5 ounces or less, and negative weights don’t make sense, the domain would be 0 < w ≤ 3.5 . Technically 0 could be included in the domain, but logically it would mean we are mailing nothing, so it doesn’t hurt to leave it out. Since possible prices are from a limited set of values, we can only define the range of this function by listing the possible values. The range is p = $0.44, $0.61, $0.78, or $0.95. Try it Now 1. The population of a small town in the year 1960 was 100 people. Since then the population has grown to 1400 people reported during the 2010 census. Choose descriptive variables for your input and output and use interval notation to write the domain and range. Notation In the previous examples, we used inequalities to describe the domain and range of the functions. This is one way to describe intervals of input and output values, but is not the only way. Let us take a moment to discuss notation for domain and range. Using inequalities, such as 0 < c ≤ 163 , 0 < w ≤ 3.5 , and 0 < h ≤ 379 imply that we are interested in all values between the low and high values, including the high values in these examples. However, occasionally we are interested in a specific list of numbers like the range for the price to send letters, p = $0.44, $0.61, $0.78, or $0.95. These numbers represent a set of specific values: {0.44, 0.61, 0.78, 0.95} Representing values as a set, or giving instructions on how a set is built, leads us to another type of notation to describe the domain and range. Suppose we want to describe the values for a variable x that are 10 or greater, but less than 30. In inequalities, we would write 10 ≤ x < 30 .
Section 1.2 Domain and Range 23 When describing domains and ranges, we sometimes extend this into set-builder notation, which would look like this: { x |10 ≤ x < 30} . The curly brackets {} are read as “the set of”, and the vertical bar | is read as “such that”, so altogether we would read { x |10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x and x is less than 30.” When describing ranges in set-builder notation, we could similarly write something like { f ( x) | 0 < f ( x) < 100} , or if the output had its own variable, we could use it. So for our tree height example above, we could write for the range {h | 0 < h ≤ 379} . In set-builder notation, if a domain or range is not limited, we could write {t | t is a real number} , or
{t | t ∈ } , read as “the set of t-values such that t is an element of the set of real numbers. A more compact alternative to set-builder notation is interval notation, in which intervals of values are referred to by the starting and ending values. Curved parentheses are used for “strictly less than,” and square brackets are used for “less than or equal to.” Since infinity is not a number, we can’t include it in the interval, so we always use curved parentheses with ∞ and -∞. The table below will help you see how inequalities correspond to set-builder notation and interval notation: Inequality 5 < h ≤ 10
Set Builder Notation {h | 5 < h ≤ 10}
Interval notation (5, 10]
5 ≤ h < 10
{h | 5 ≤ h < 10} {h | 5 < h < 10} {h | h < 10} {h | h ≥ 10} {h | h ∈ }
[5, 10)
5 < h < 10
h < 10 h ≥ 10
all real numbers
(5, 10) (−∞,10) [10, ∞) (−∞, ∞)
To combine two intervals together, using inequalities or set-builder notation we can use the word “or”. In interval notation, we use the union symbol, ∪ , to combine two unconnected intervals together. Example 3 Describe the intervals of values shown on the line graph below using set builder and interval notations.
24 Chapter 1 To describe the values, x, that lie in the intervals shown above we would say, “x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” As an inequality it is: 1 ≤ x ≤ 3 or x > 5
In set builder notation: { x |1 ≤ x ≤ 3 or x > 5} In interval notation: [1,3] ∪ (5, ∞) Remember when writing or reading interval notation: Using a square bracket [ means the start value is included in the set Using a parenthesis ( means the start value is not included in the set Try it Now 2. Given the following interval, write its meaning in words, set builder notation, and interval notation.
Domain and Range from Graphs We can also talk about domain and range based on graphs. Since domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the graph. Remember that input values are almost always shown along the horizontal axis of the graph. Likewise, since range is the set of possible output values, the range of a graph we can see from the possible values along the vertical axis of the graph. Be careful – if the graph continues beyond the window on which we can see the graph, the domain and range might be larger than the values we can see.
Section 1.2 Domain and Range 25 Example 4 Determine the domain and range of the graph below.
In the graph above 3, the input quantity along the horizontal axis appears to be “year”, which we could notate with the variable y. The output is “thousands of barrels of oil per day”, which we might notate with the variable b, for barrels. The graph would likely continue to the left and right beyond what is shown, but based on the portion of the graph that is shown to us, we can determine the domain is 1975 ≤ y ≤ 2008 , and the range is approximately 180 ≤ b ≤ 2010 . In interval notation, the domain would be [1975, 2008] and the range would be about [180, 2010]. For the range, we have to approximate the smallest and largest outputs since they don’t fall exactly on the grid lines. Remember that, as in the previous example, x and y are not always the input and output variables. Using descriptive variables is an important tool to remembering the context of the problem.
3
http://commons.wikimedia.org/wiki/File:Alaska_Crude_Oil_Production.PNG, CC-BY-SA, July 19, 2010
26 Chapter 1 Try it Now 3. Given the graph below write the domain and range in interval notation
Domains and Ranges of the Toolkit functions We will now return to our set of toolkit functions to note the domain and range of each. Constant Function: f ( x) = c The domain here is not restricted; x can be anything. When this is the case we say the domain is all real numbers. The outputs are limited to the constant value of the function. Domain: (−∞, ∞) Range: [c] Since there is only one output value, we list it by itself in square brackets. Identity Function: f ( x) = x Domain: (−∞, ∞) Range: (−∞, ∞) Quadratic Function: f ( x) = x 2 Domain: (−∞, ∞) Range: [0, ∞) Multiplying a negative or positive number by itself can only yield a positive output. Cubic Function: f ( x) = x 3 Domain: (−∞, ∞) Range: (−∞, ∞)
Section 1.2 Domain and Range 27 1 x Domain: (−∞, 0) ∪ (0, ∞) Range: (−∞, 0) ∪ (0, ∞) We cannot divide by 0 so we must exclude 0 from the domain. One divide by any value can never be 0, so the range will not include 0.
Reciprocal: f ( x) =
Reciprocal squared: f ( x) =
1 x2
Domain: (−∞, 0) ∪ (0, ∞) Range: (0, ∞) We cannot divide by 0 so we must exclude 0 from the domain. Cube Root: f ( x) = 3 x Domain: (−∞, ∞) Range: (−∞, ∞) Square Root: f ( x) = 2 x , commonly just written as, f ( x) = x Domain: [0, ∞) Range: [0, ∞) When dealing with the set of real numbers we cannot take the square root of a negative number so the domain is limited to 0 or greater. Absolute Value Function: f ( x) = x Domain: (−∞, ∞) Range: [0, ∞) Since absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Example 4.5 Find the domain of each function: a) f ( x) = 2 x + 4
b) g ( x) =
3 6 − 3x
a) Since we cannot take the square root of a negative number, we need the inside of the square root to be non-negative. x + 4 ≥ 0 when x ≥ −4 . The domain of f(x) is [−4, ∞) . b) We cannot divide by zero, so we need the denominator to be non-zero. 6 − 3 x = 0 when x = 2, so we must exclude 2 from the domain. The domain of g(x) is (−∞,2) ∪ (2, ∞) .
28 Chapter 1 Piecewise Functions In the toolkit functions we introduced the absolute value function f ( x) = x . With a domain of all real numbers and a range of values greater than or equal to 0, the absolute value can be defined as the magnitude or modulus of a number, a real number value regardless of sign, the size of the number, or the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value the output is unchanged f ( x) = x if x ≥ 0 If we input a negative value the sign must change from negative to positive. since multiplying a negative value by -1 makes it positive. f ( x) = − x if x < 0 Since this requires two different processes or pieces, the absolute value function is often called the most basic piecewise defined function. Piecewise Function A piecewise function is a function in which the formula used depends upon the domain the input lies in. We notate this idea like:
formula 1 if f ( x) = formula 2 if formula 3 if
domain to use formula 1 domain to use formula 2 domain to use formula 3
Example 5 A museum charges $5 per person for a guided tour with a group of 1 to 9 people, or a fixed $50 fee for 10 or more people in the group. Set up a function relating the number of people, n, to the cost, C. To set up this function, two different formulas would be needed. C = 5n would work for n values under 10, and C = 50 would work for values of n ten or greater. Notating this: 5n if 0 < n < 10 C ( n) = n ≥ 10 50 if Example 6 A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. if 25 C(g) = 25 + 10( g − 2) if
0< g 2
Since each of the component functions are from our library of Toolkit functions, we know their shapes. We can imagine graphing each function, then limiting the graph to the indicated domain. At the endpoints of the domain, we put open circles to indicate where the endpoint is not included, due to a strictly-less-than inequality, and a closed circle where the endpoint is included, due to a less-than-or-equal-to inequality.
Now that we have each piece individually, we combine them onto the same graph:
30 Chapter 1 Try it Now 4. At Pierce College during the 2009-2010 school year tuition rates for in-state residents were $89.50 per credit for the first 10 credits, $33 per credit for credits 11-18, and for over 18 credits the rate is $73 per credit 4. Write a piecewise defined function for the total tuition, T, at Pierce College during 2009-2010 as a function of the number of credits taken, c. Be sure to consider a reasonable domain and range. Important Topics of this Section Definition of domain Definition of range Inequalities Interval notation Set builder notation Domain and Range from graphs Domain and Range of toolkit functions Piecewise defined functions Try it Now Answers 1. Domain; y = years [1960,2010] ; Range, p = population, [100,1400] 2. a. Values that are less than or equal to -2, or values that are greater than or equal to 1 and less than 3 b. { x | x ≤ −2 or − 1 ≤ x < 3} c. (−∞, −2] ∪ [−1,3) 3. Domain; y=years, [1952,2002] ; Range, p=population in millions, [40,88]
89.5c if 4. T (c) = 895 + 33(c − 10) if 1159 + 73(c − 18) if
c ≤ 10 10 < c ≤ 18 Tuition, T, as a function of credits, c. c > 18
Reasonable domain should be whole numbers 0 to (answers may vary), e.g. [0, 23] Reasonable range should be $0 – (answers may vary), e.g. [0,1524]
4
https://www.pierce.ctc.edu/dist/tuition/ref/files/0910_tuition_rate.pdf, retrieved August 6, 2010
Section 1.2 Domain and Range 31
Section 1.2 Exercises Write the domain and range of the function using interval notation.
1.
2.
Write the domain and range of each graph as an inequality.
3.
4.
Suppose that you are holding your toy submarine under the water. You release it and it begins to ascend. The graph models the depth of the submarine as a function of time, stopping once the sub surfaces. What is the domain and range of the function in the graph?
5.
6.
32 Chapter 1 Find the domain of each function 7. f (= x) 3 x − 2
8. f (= x) 5 x + 3
9. f ( x ) =3 − 6 − 2 x
10. f ( x ) = 5 − 10 − 2 x
11. f ( x ) =
9 x 6 −
12. f ( x ) =
6 x 8 −
13. f ( x ) =
3x + 1 4x + 2
14. f ( x ) =
5x + 3 4x −1
15. f ( x ) =
x+4 x−4
16. f ( x ) =
x+5 x−6
17. f ( x ) =
x 3 − x + 9 x −22
18. f ( x ) =
2
x 8 − x + 8 x −9 2
Given each function, evaluate: f (−1) , f (0) , f (2) , f (4) 4 x − 9 if 7 x + 3 if x < 0 20. f ( x ) = 19. f ( x ) = 4 x − 18 if 7 x + 6 if x ≥ 0 x 2 − 2 21. f ( x ) = 4 + x − 5
if if
5 x if 23. = f ( x ) 3 if x 2 if
x 1, then the graph will be stretched If 0< k < 1, then the graph will be compressed If k < 0, then there will be combination of a vertical stretch or compression with a vertical reflection
Section 1.5 Transformation of Functions 73 Example 14 A function f(x) is given as a table below. Create a table for the function g ( x) = x f(x)
2 1
4 3
6 7
1 f ( x) 2
8 11
1 f ( x) tells us that the output values of g are half of the output 2 values of f with the same inputs. For example, we know that f (4) = 3 . Then 3 1 1 g (4) = f (4) = (3) = 2 2 2
The formula g ( x) =
x g(x)
2 1/2
4 3/2
6 7/2
8 11/2
The result is that the function g(x) has been compressed vertically by ½. Each output value has been cut in half, so the graph would now be half the original height. Example 15 The graph to the right is a transformation of the toolkit function f ( x) = x 3 . Relate this new function g(x) to f(x), then find a formula for g(x). When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g (2) = 2 . With the basic cubic function at the same input, f (2) = 2 3 = 8 . Based on that, it appears that the outputs of g are ¼ the outputs of the function f, 1 since g (2) = f (2) . From this we can fairly safely 4 conclude that: 1 g ( x) = f ( x) 4 We can write a formula for g by using the definition of the function f 1 1 g ( x) = f ( x) = x 3 4 4 Now we consider changes to the inside of a function.
74 Chapter 1 Example 16 Returning to the fruit fly population we looked at earlier, suppose the scientist is now comparing it to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in 1 hour the same amount the original population did in 2 hours, and in 2 hours, will progress as much as the original population did in 4 hours. Sketch a graph of this population. Symbolically, we could write R (1) = P (2) R(2) = P(4) , and in general, R(t ) = P(2t ) Graphing this, Original population, P(t)
Transformed, R(t)
Note the effect on the graph is a horizontal compression, where all input values are half their original distance from the vertical axis. Horizontal Stretch/Compression Given a function f(x), if we define a new function g(x) as g ( x) = f (kx) , where k is a constant then g(x) is a horizontal stretch or compression of the function f(x). If k > 1, then the graph will be compressed by 1
k 1 If 0< k < 1, then the graph will be stretched by k If k < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection.
Section 1.5 Transformation of Functions 75 Example 17
1 A function f(x) is given as a table below. Create a table for the function g ( x) = f x 2
x f(x)
2 1
4 3
6 7
8 11
1 The formula g ( x) = f x tells us that the output values for g are the same as the 2 output values for the function f at an input half the size. Notice that we don’t have 1 enough information to determine g (2) since g (2) = f ⋅ 2 = f (1) , and we do not 2 have a value for f (1) in our table. Our input values to g will need to be twice as large to get inputs for f that we can evaluate. For example, we can determine g (4) since 1 g (4) = f ⋅ 4 = f (2) = 1 . 2
x g(x)
4 1
8 3
12 7
16 11
Since each input value has been doubled, the result is that the function g(x) has been stretched horizontally by 2. Example 18 Two graphs are shown below. Relate the function g(x) to f(x) f(x)
g(x)
The graph of g(x) looks like the graph of f(x) horizontally compressed. Since f(x) ends at (6,4) and g(x) ends at (2,4) we can see that the x values have been compressed by 1/3, because 6(1/3) = 2. We might also notice that g (2) = f (6 ) , and g (1) = f (3) . Either way, we can describe this relationship as g ( x) = f (3 x ) . This is a horizontal compression by 1/3.
76 Chapter 1 Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by 4, we 1 need a coefficient of 1/4 in our function: f x . This means the input values must be 4 four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. Try it Now 4. Write a formula for the toolkit square root function horizontally stretched by three. It is useful to note that for most toolkit functions, a horizontal stretch or vertical stretch can be represented in other ways. For example, a horizontal compression of the function 2 f ( x ) = x 2 by ½ would result in a new function g ( x ) = ( 2 x ) , but this can also be written as g ( x ) = 4 x 2 , a vertical stretch of f(x) by 4. When writing a formula for a transformed toolkit, we only need to find one transformation that would produce the graph. Combining Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3. When we see an expression like 2 f ( x) + 3 , which transformation should we start with? The answer here follows nicely from order of operations, for outside transformations. Given the output value of f(x), we first multiply by 2, causing the vertical stretch, then add 3, causing the vertical shift. (Multiplication before Addition) Combining Vertical Transformations When combining vertical transformations written in the form af ( x) + k , first vertically stretch by a, then vertically shift by k. Horizontal transformations are a little trickier to think about. When we write g ( x) = f (2 x + 3) for example, we have to think about how the inputs to the g function relate to the inputs to the f function. Suppose we know f (7) = 12 . What input to g would produce that output? In other words, what value of x will allow g ( x) = f (2 x + 3) = f (12) ? We would need 2 x + 3 = 12 . To solve for x, we would first subtract 3, resulting in horizontal shift, then divide by 2, causing a horizontal compression.
Section 1.5 Transformation of Functions 77 Combining Horizontal Transformations When combining horizontal transformations written in the form f (bx + p ) , first horizontally shift by p, then horizontally stretch by 1/b. This format ends up being very difficult to work with, since it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function. p f (bx + p ) = f b x + b Factoring in this way allows us to horizontally stretch first, then shift horizontally. Combining Horizontal Transformations (Factored Form) When combining horizontal transformations written in the form f (b( x + h)) , first horizontally stretch by 1/b, then horizontally shift by h. Independence of Horizontal and Vertical Transformations Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are done first. Example 19 Given the table of values for the function f(x) below, create a table of values for the function g ( x) = 2 f (3 x) + 1 x f(x)
6 10
12 14
18 15
24 17
There are 3 steps to this transformation and we will work from the inside out. Starting with the horizontal transformations, f (3 x) is a horizontal compression by 1/3, which means we multiply each x value by 1/3. x f (3 x)
2 10
4 14
6 15
8 17
Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. x 2 f (3 x)
2 20
4 28
6 30
8 34
Finally, we can apply the vertical shift, which will add 1 to all the output values. x 2 4 6 8 g ( x) = 2 f (3 x) + 1 21 29 31 35
78 Chapter 1 Example 20
1 Using the graph of f(x) below, sketch a graph of k ( x) = f x + 1 − 3 2
To make things simpler, we’ll start by factoring out the inside of the function 1 1 f x + 1 − 3 = f ( x + 2) − 3 2 2 By factoring the inside, we can first horizontally stretch by 2, as indicated by the ½ on the inside of the function. Remember twice the size of 0 is still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch to (4,0). Next, we horizontally shift left by 2 units, as indicated by the x+2. Last, we vertically shift down by 3 to complete our sketch, as indicated by the -3 on the outside of the function. Horizontal stretch by 2
Horizontal shift left by 2
Vertical shift down 3
Section 1.5 Transformation of Functions 79 Example 21 Write an equation for the transformed graph of the quadratic function shown. Since this is a quadratic function, first consider what the basic quadratic tool kit function looks like and how this has changed. Observing the graph, we notice several transformations: The original tool kit function has been flipped over the x axis, some kind of stretch or compression has occurred, and we can see a shift to the right 3 units and a shift up 1 unit. In total there are four operations: Vertical reflection, requiring a negative sign outside the function Vertical Stretch or Horizontal Compression* Horizontal Shift Right 3 units, which tells us to put x-3 on the inside of the function Vertical Shift up 1 unit, telling us to add 1 on the outside of the function *
It is unclear from the graph whether it is showing a vertical stretch or a horizontal compression. For the quadratic, it turns out we could represent it either way, so we’ll use a vertical stretch. You may be able to determine the vertical stretch by observation. By observation, the basic tool kit function has a vertex at (0, 0) and symmetrical points at (1, 1) and (-1, 1). These points are one unit up and one unit over from the vertex. The new points on the transformed graph are one unit away horizontally but 2 units away vertically. They have been stretched vertically by two. Not everyone can see this by simply looking at the graph. If you can, great, but if not, we can solve for it. First, we will write the equation for this graph, with an unknown vertical stretch. f ( x) = x 2 − f ( x) = − x 2 − af ( x) = −ax 2 − af ( x − 3) = −a ( x − 3) 2 − af ( x − 3) + 1 = −a ( x − 3) 2 + 1
The original function Vertically reflected Vertically stretched Shifted right 3 Shifted up 1
We now know our graph is going to have an equation of the form g ( x) = −a ( x − 3) 2 + 1 . To find the vertical stretch, we can identify any point on the graph (other than the highest point), such as the point (2,-1), which tells us g (2) = −1 . Using our general formula, and substituting 2 for x, and -1 for g(x)
80 Chapter 1 − 1 = − a (2 − 3) 2 + 1 − 1 = −a + 1 − 2 = −a 2=a
This tells us that to produce the graph we need a vertical stretch by two. The function that produces this graph is therefore g ( x) = −2( x − 3) 2 + 1 . Try it Now 5. Consider the linear function g ( x) = −2 x + 1 . Describe its transformation in words using the identity tool kit function f(x) = x as a reference. Example 22 On what interval(s) is the function g ( x) =
−2
(x − 1)2
+ 3 increasing and decreasing?
This is a transformation of the toolkit reciprocal squared function, f ( x) = − 2 f ( x) =
−2 x2
− 2 f ( x − 1) =
1 : x2
A vertical flip and vertical stretch by 2 −2
A shift right by 1
(x − 1)2
− 2 f ( x − 1) + 3 =
−2
(x − 1)2
+3
A shift up by 3
The basic reciprocal squared function is increasing on (−∞,0) and decreasing on (0, ∞) . Because of the vertical flip, the g(x) function will be decreasing on the left and increasing on the right. The horizontal shift right by 1 will also shift these intervals to the right one. From this, we can determine g(x) will be increasing on (1, ∞) and decreasing on (−∞,1) . We also could graph the transformation to help us determine these intervals. Try it Now 6. On what interval(s) is the function h(t ) = (t − 3) 3 + 2 concave up and down?
Section 1.5 Transformation of Functions 81 Important Topics of This Section Transformations Vertical Shift (up & down) Horizontal Shifts (left & right) Reflections over the vertical & horizontal axis Even & Odd functions Vertical Stretches & Compressions Horizontal Stretches & Compressions Combinations of Transformation Try it Now Answers 1. b(t ) = h(t ) + 10 = −4.9t 2 + 30t + 10 2. a. Horizontal shift b. The function is shifted to the LEFT by 2 units. c. Shown to the right
3. Shown to the right Notice: g(x) = f(-x) looks the same as f(x) 1 4. g ( x) = f x so using the square root function we get 3 1 g ( x) = x 3 5. The identity tool kit function f(x) = x has been transformed in 3 steps a. Vertically stretched by 2. b. Vertically reflected over the x axis. c. Vertically shifted up by 1 unit. 6. h(t) is concave down on (−∞,3) and concave up on (3, ∞)
82 Chapter 1
Section 1.5 Exercises Describe how each function is a transformation of the original function f ( x) 1. f ( x − 49 ) 2. f ( x + 43) 3. f ( x + 3)
4. f ( x − 4)
7. f ( x ) − 2
8. f ( x ) − 7
9. f ( x − 2 ) + 3
10. f ( x + 4 ) − 1
5. f ( x ) + 5
6. f ( x ) + 8
11. Write a formula for f ( x) = x shifted up 1 unit and left 2 units. 12. Write a formula for f ( x) = x shifted down 3 units and right 1 unit. 1 shifted down 4 units and right 3 units. x 1 14. Write a formula for f ( x) = 2 shifted up 2 units and left 4 units. x 13. Write a formula for f ( x) =
15. Tables of values for f ( x) , g ( x) , and h( x) are given below. Write g ( x) and h( x) as transformations of f ( x) . x -2 -1 0 1 2 f(x) -2 -1 -3 1 2
x -1 0 1 2 3 g(x) -2 -1 -3 1 2
-2 -1 0 1 2 x h(x) -1 0 -2 2 3
16. Tables of values for f ( x) , g ( x) , and h( x) are given below. Write g ( x) and h( x) as transformations of f ( x) . x -2 -1 0 1 2 f(x) -1 -3 4 2 1
x -3 -2 -1 0 1 g(x) -1 -3 4 2 1
-2 -1 0 1 2 x h(x) -2 -4 3 1 0
The graph of f ( x ) = 2 x is shown. Sketch a graph of each transformation of f ( x) 17. g ( x= ) 2x + 1 18. h ( x= ) 2x − 3 19. w ( x ) = 2 x −1 20. q ( x ) = 2 x +3
Section 1.5 Transformation of Functions 83 Sketch a graph of each function as a transformation of a toolkit function. 21. f ( t ) =(t + 1) 2 − 3 22. h ( x ) = x − 1 + 4 23. k ( x ) =( x − 2 ) − 1 3
24. m ( t ) =3 + t + 2 Write an equation for each function graphed below.
25.
26.
27.
28.
Find a formula for each of the transformations of the square root whose graphs are given below.
29.
30.
84 Chapter 1 The graph of f ( x ) = 2 x is shown. Sketch a graph of each transformation of f ( x) 31. g ( x ) = −2 x + 1 32. h ( x ) = 2− x
33. Starting with the graph of f ( x ) = 6 x write the equation of the graph that results from a. reflecting f ( x) about the x-axis and the y-axis b. reflecting f ( x) about the x-axis, shifting left 2 units, and down 3 units 34. Starting with the graph of f ( x ) = 4 x write the equation of the graph that results from a. reflecting f ( x) about the x-axis b. reflecting f ( x) about the y-axis, shifting right 4 units, and up 2 units Write an equation for each function graphed below.
35.
36.
37.
38.
Section 1.5 Transformation of Functions 85 39. For each equation below, determine if the function is Odd, Even, or Neither. a. f ( x ) = 3 x 4 b. g ( x) = x 1 c. h ( x )= + 3 x x 40. For each equation below, determine if the function is Odd, Even, or Neither. 2 a. f ( x= ) ( x − 2) b. g ( x ) = 2 x 4 c. h ( x= ) 2 x − x3 Describe how each function is a transformation of the original function f ( x) . 41. − f ( x) 42. f (− x) 44. 6 f ( x) 43. 4 f ( x) 45. f (5 x) 46. f (2 x) 1 1 47. f x 48. f x 3 5 50. − f (3 x) 49. 3 f ( − x ) Write a formula for the function that results when the given toolkit function is transformed as described. 51. f ( x) = x reflected over the y axis and horizontally compressed by a factor of
1 . 4
52. f ( x) = x reflected over the x axis and horizontally stretched by a factor of 2. 1 1 vertically compressed by a factor of , then shifted to the left 2 units and 2 3 x down 3 units.
53. f ( x) =
54. f ( x) = 2 units.
1 vertically stretched by a factor of 8, then shifted to the right 4 units and up x
55. f ( x) = x 2 horizontally compressed by a factor of and up 1 unit.
1 , then shifted to the right 5 units 2
56. f ( x) = x 2 horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units.
86 Chapter 1 Describe how each formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 2 2 57. f ( x ) = 4 ( x + 1) − 5 58. g ( x) = 5 ( x + 3) − 2
−2 x − 4 + 3 59. h ( x ) = 61. m ( x ) =
1 3 x 2
60. k ( x ) = −3 x − 1 62. n (= x)
2
1 x−2 3 3
1 63. = p ( x) x − 3 3
1 64. = q ( x) x +1 4
65. a ( x ) =
66. b(x ) = 3 − x − 6
−x + 4
Determine the interval(s) on which the function is increasing and decreasing. 67. f ( x ) = 4 ( x + 1) − 5
68. g ( x) = 5 ( x + 3) − 2
69. a ( x ) =
70. k ( x ) = −3 x − 1
2
−x + 4
2
Determine the interval(s) on which the function is concave up and concave down. 71. m( x ) = −2( x + 3) 3 + 1 2
1 73. = p ( x) x − 3 3
72. b ( x ) =
3
−x − 6
74. k ( x ) = −3 x − 1
Section 1.5 Transformation of Functions 87 The function f ( x) is graphed here. Write an equation for each graph below as a transformation of f ( x) .
75.
76.
77.
78.
79.
80.
81.
82.
83.
84.
85.
86.
88 Chapter 1 Write an equation for each transformed toolkit function graphed below.
87.
88.
89.
90.
91.
92.
93.
94.
95.
96.
97.
98.
Section 1.5 Transformation of Functions 89 99. Suppose you have a function y = f ( x) such that the domain of f ( x) is 1 ≤ x ≤ 6 and the range of f ( x) is −3 ≤ y ≤ 5. [UW] a. What is the domain of (2( ? f x − 3)) b. What is the range of f (2( x − 3)) ? c. What is the domain of 2 f ( x) − 3 ? d. What is the range of 2 f ( x) − 3 ? e. Can you find constants B and C so that the domain of f ( B( x − C )) is 8 ≤ x ≤ 9? f. Can you find constants A and D so that the range of Af ( x) + D is 0 ≤ y ≤ 1?
90 Chapter 1
Section 1.6 Inverse Functions A fashion designer is travelling to Milan for a fashion show. He asks his assistant, Betty, what 75 degrees Fahrenheit is in Celsius, and after a quick search on Google, she finds 5 5 the formula C = ( F − 32) . Using this formula, she calculates (75 − 32) ≈ 24 degrees 9 9 Celsius. The next day, the designer sends his assistant the week’s weather forecast for Milan, and asks her to convert the temperatures to Fahrenheit.
At first, Betty might consider using the formula she has already found to do the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write: 5 26 = ( F − 32) 9 9 26 ⋅ = F − 32 5 9 F = 26 ⋅ + 32 ≈ 79 5 After considering this option for a moment, she realizes that solving the equation for each of the temperatures would get awfully tedious, and realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one which takes the Celsius temperature and outputs the Fahrenheit temperature. This is the idea of an inverse function, where the input becomes the output and the output becomes the input. Inverse Function If f (a ) = b , then a function g(x) is an inverse of f if g (b) = a . The inverse of f(x) is typically notated f −1 ( x) , which is read “f inverse of x”, so equivalently, if f (a ) = b then f −1 (b) = a . Important: The raised -1 used in the notation for inverse functions is simply a notation, and does not designate an exponent or power of -1.
Section 1.6 Inverse Functions 91 Example 1 If for a particular function, f (2) = 4 , what do we know about the inverse? The inverse function reverses which quantity is input and which quantity is output, so if f (2) = 4 , then f −1 (4) = 2 . Alternatively, if you want to re-name the inverse function g(x), then g(4) = 2 Try it Now 1. Given that h −1 (6) = 2 , what do we know about the original function h(x)? Notice that original function and the inverse function undo each other. If f (a ) = b , then f −1 (b) = a , returning us to the original input. More simply put, if you compose these functions together you get the original input as your answer. f −1 ( f (a ) ) = a and f ( f −1 (b) ) = b Domain of f
f (x)
a
Range of f b
f
−1
( x)
Since the outputs of the function f are the inputs to f −1 , the range of f is also the domain of f −1 . Likewise, since the inputs to f are the outputs of f −1 , the domain of f is the range of f −1 . Basically, like how the input and output values switch, the domain & ranges switch as well. But be careful, because sometimes a function doesn’t even have an inverse function, or only has an inverse on a limited domain. For example, the inverse of f ( x) = x is f −1 ( x) = x 2 , since a square “undoes” a square root, but it is only the inverse of f(x) on the domain [0,∞), since that is the range of f ( x) = x . Example 2 The function f ( x) = 2 x has domain (−∞, ∞) and range (0, ∞) , what would we expect the domain and range of f −1 to be? We would expect f −1 to swap the domain and range of f, so f domain (0, ∞) and range (−∞, ∞) .
−1
would have
92 Chapter 1 Example 3 A function f(t) is given as a table below, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1 (70) t (minutes) f(t) (miles)
30 20
50 40
70 60
90 70
The inverse function takes an output of f and returns an input for f. So in the expression f −1 (70) , the 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so f −1 (70) = 90 . Interpreting this, it means that to drive 70 miles, it took 90 minutes. Alternatively, recall the definition of the inverse was that if f (a ) = b then f −1 (b) = a . By this definition, if you are given f −1 (70) = a then you are looking for a value a so that f (a ) = 70 . In this case, we are looking for a t so that f (t ) = 70 , which is when t = 90. Try it Now 2. Using the table below t (minutes) f(t) (miles)
30 20
50 40
60 50
70 60
90 70
Find and interpret the following a. f (60) b. f −1 (60) Example 4 A function g(x) is given as a graph below. Find g (3) and g −1 (3)
To evaluate g (3) , we find 3 on the horizontal axis and find the corresponding output value on the vertical axis. The point (3, 1) tells us that g (3) = 1
Section 1.6 Inverse Functions 93 To evaluate g −1 (3) , recall that by definition g −1 (3) means g(x) = 3. By looking for the output value 3 on the vertical axis we find the point (5, 3) on the graph, which means g(5) = 3, so by definition g −1 (3) = 5 . Try it Now 3. Using the graph in Example 4 above a. find g −1 (1) b. estimate g −1 (4) Example 5 Returning to our designer’s assistant, find a formula for the inverse function that gives Fahrenheit temperature given a Celsius temperature. A quick Google search would find the inverse function, but alternatively, Betty might look back at how she solved for the Fahrenheit temperature for a specific Celsius value, and repeat the process in general 5 C = ( F − 32) 9 9 C ⋅ = F − 32 5 9 F = C + 32 5 By solving in general, we have uncovered the inverse function. If 5 C = h( F ) = ( F − 32) 9 Then 9 F = h −1 (C ) = C + 32 5 In this case, we introduced a function h to represent the conversion since the input and output variables are descriptive, and writing C −1 could get confusing. It is important to note that not all functions will have an inverse function. Since the inverse f −1 ( x) takes an output of f and returns an input of f, in order for f −1 to itself be a function, then each output of f (input to f −1 ) must correspond to exactly one input of f (output of f −1 ) in order for f −1 to be a function. You might recall that this is the definition of a one-to-one function.
94 Chapter 1 Properties of Inverses In order for a function to have an inverse, it must be a one-to-one function. In some cases, it is desirable to have an inverse for a function even though the function is not one-to-one. In those cases, we can often limit the domain of the original function to an interval on which the function is one-to-one, then find an inverse only on that interval. If you have not already done so, go back to the toolkit functions that were not one-to-one and limit or restrict the domain of the original function so that it is one-to-one. If you are not sure how to do this, proceed to Example 6. Example 6 The quadratic function h( x) = x 2 is not one-to-one. Find a domain on which this function is one-to-one, and find the inverse on that domain. We can limit the domain to [0, ∞) to restrict the graph to a portion that is one-to-one, and find an inverse on this limited domain. You may have already guessed that since we undo a square with a square root, the inverse of h( x) = x 2 on this domain is h −1 ( x) = x . You can also solve for the inverse function algebraically. If h( x) = x 2 , we can introduce the variable y to represent the output values, allowing us to write y = x 2 . To find the inverse we solve for the input variable To solve for x we take the square root of each side.
y = x 2 and get
y = x , so
x = ± y . We have restricted x to being non-negative, so we’ll use the positive square root, x =
y or h −1 ( y ) =
y . In cases like this where the variables are not descriptive,
it is common to see the inverse function rewritten with the variable x: h −1 ( x) = x . Rewriting the inverse using the variable x is often required for graphing inverse functions using calculators or computers. Note that the domain and range of the square root function do correspond with the range and domain of the quadratic function on the limited domain. In fact, if we graph h(x) on the restricted domain and h −1 ( x) on the same axes, we can notice symmetry: the graph of h −1 ( x) is the graph of h(x) reflected over the line y = x.
h(x)
y=x
h −1 ( x)
Section 1.6 Inverse Functions 95 Example 7 Given the graph of f(x) shown, sketch a graph of f
−1
( x) .
This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0,∞) and range of (-∞,∞), so the inverse will have a domain of (-∞,∞) and range of (0,∞). Reflecting this graph of the line y = x, the point (1, 0) reflects to (0, 1), and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph: f
−1
( x)
f (x)
Important Topics of this Section Definition of an inverse function Composition of inverse functions yield the original input value Not every function has an inverse function To have an inverse a function must be one-to-one Restricting the domain of functions that are not one-to-one. Try it Now Answers 1. g (2) = 6 2.a. f (60) = 50 . In 60 minutes, 50 miles are traveled. b. f −1 (60) = 70 . To travel 60 miles, it will take 70 minutes. 3. a. g −1 (1) = 3 b. g −1 (4) = 5.5 (this is an approximation – answers may vary slightly)
96 Chapter 1
Section 1.6 Exercises Assume that the function f is a one-to-one function. 1. If f (6) = 7 , find f −1 (7) 2. If f (3) = 2 , find f −1 (2) −8 , find f (−8) −1 , find f (−1) 4. If f −1 ( −2 ) = 3. If f −1 ( −4 ) = 5. If f ( 5 ) = 2 , find ( f ( 5 ) )
6. If f (1) = 4 , find ( f (1) )
−1
−1
7. Using the graph of f ( x) shown a. Find f ( 0 ) b. Solve f ( x) = 0 c. Find f −1 ( 0 ) d. Solve f −1 ( x ) = 0
8. Using the graph shown a. Find g (1) b. Solve g ( x) = 1 c. Find g −1 (1) d. Solve g −1 ( x ) = 1
9. Use the table below to find the indicated quantities. x f(x)
0 8
1 0
2 7
a. Find f (1) b. Solve f ( x) = 3 c. Find f −1 ( 0 ) d. Solve f −1 ( x ) = 7
3 4
4 2
5 6
6 5
7 3
8 9
9 1
Section 1.6 Inverse Functions 97 10. Use the table below to fill in the missing values. t h(t)
0 6
1 0
2 1
3 7
4 2
5 3
6 5
7 4
8 9
a. Find h ( 6 ) b. Solve h(t ) = 0 c. Find h −1 ( 5 ) d. Solve h −1 ( t ) = 1 For each table below, create a table for f −1 ( x ) . 11. x 12. 3 6 9 13 14 f(x) 1 4 7 12 16
3 5 7 13 15 x f(x) 2 6 9 11 16
For each function below, find f −1 ( x) 13. f ( x )= x + 3
14. f ( x )= x + 5
15. f ( x ) = 2 – x
16. f ( x )= 3 − x
17. f (= x ) 11x + 7
18. f ( x )= 9 + 10 x
For each function, find a domain on which f is one-to-one and non-decreasing, then find the inverse of f restricted to that domain. 2 2 19. f ( x = 20. f ( x= ) 7 (x+ ) ) ( x − 6) 21. f ( x= ) x2 − 5 23. If f ( x= x) ) x3 − 5 and g ( =
22. f ( x= ) x2 + 1 3
x + 5 , find
a. ( f g ( x)) b.
g ( f ( x))
c. What does this tell us about the relationship between f ( x) and g ( x) ? 2x x and g ( x) = , find 1− x 2+ x a. ( f g ( x))
24. If f ( x) = b.
g ( f ( x))
c. What does this tell us about the relationship between f ( x) and g ( x) ?
98 Chapter 1
Chapter 2: Linear Functions
Chapter one was a window that gave us a peek into the entire course. Our goal was to understand the basic structure of functions and function notation, the toolkit functions, domain and range, how to recognize and understand composition and transformations of functions and how to understand and utilize inverse functions. With these basic components in hand we will further research the specific details and intricacies of each type of function in our toolkit and use them to model the world around us. Mathematical Modeling As we approach day to day life we often need to quantify the things around us, giving structure and numeric value to various situations. This ability to add structure enables us to make choices based on patterns we see that are weighted and systematic. With this structure in place we can model and even predict behavior to make decisions. Adding a numerical structure to a real world situation is called Mathematical Modeling. When modeling real world scenarios, there are some common growth patterns that are regularly observed. We will devote this chapter and the rest of the book to the study of the functions used to model these growth patterns. Section 2.1 Linear Functions ........................................................................................ 99 Section 2.2 Graphs of Linear Functions ..................................................................... 111 Section 2.3 Modeling with Linear Functions .............................................................. 126 Section 2.4 Fitting Linear Models to Data .................................................................. 138 Section 2.5 Absolute Value Functions ........................................................................ 146
Section 2.1 Linear Functions As you hop into a taxicab in Las Vegas, the meter will immediately read $3.30; this is the “drop” charge made when the taximeter is activated. After that initial fee, the taximeter will add $2.40 for each mile the taxi drives 1. In this scenario, the total taxi fare depends upon the number of miles ridden in the taxi, and we can ask whether it is possible to model this type of scenario with a function. Using descriptive variables, we choose m for miles and C for Cost in dollars as a function of miles: C(m). We know for certain that C (0) = 3.30 , since the $3.30 drop charge is assessed regardless of how many miles are driven. Since $2.40 is added for each mile driven, then C (1) = 3.30 + 2.40 = 5.70 If we then drove a second mile, another $2.40 would be added to the cost: C (2) = 3.30 + 2.40 + 2.40 = 3.30 + 2.40(2) = 8.10 1
http://taxi.state.nv.us/FaresFees.htm, retrieved July 28, 2010. There is also a waiting fee assessed when the taxi is waiting at red lights, but we’ll ignore that in this discussion. This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
100 Chapter 2 If we drove a third mile, another $2.40 would be added to the cost: C (3) = 3.30 + 2.40 + 2.40 + 2.40 = 3.30 + 2.40(3) = 10.50 From this we might observe the pattern, and conclude that if m miles are driven, C (m) = 3.30 + 2.40m because we start with a $3.30 drop fee and then for each mile increase we add $2.40. It is good to verify that the units make sense in this equation. The $3.30 drop charge is measured in dollars; the $2.40 charge is measured in dollars per mile. So dollars C (m) = 3.30dollars + 2.40 (m miles ) mile When dollars per mile are multiplied by a number of miles, the result is a number of dollars, matching the units on the 3.30, and matching the desired units for the C function. Notice this equation C (m) = 3.30 + 2.40m consisted of two quantities. The first is the fixed $3.30 charge which does not change based on the value of the input. The second is the $2.40 dollars per mile value, which is a rate of change. In the equation this rate of change is multiplied by the input value. Looking at this same problem in table format we can also see the cost changes by $2.40 for every 1 mile increase. m C(m)
0 3.30
1 5.70
2 8.10
3 10.50
It is important here to note that in this equation, the rate of change is constant; over any interval, the rate of change is the same. Graphing this equation, C (m) = 3.30 + 2.40m we see the shape is a line, which is how these functions get their name: linear functions When the number of miles is zero the cost is $3.30, giving the point (0, 3.30) on the graph. This is the vertical or C(m) intercept. The graph is increasing in a straight line from left to right because for each mile the cost goes up by $2.40; this rate remains consistent. In this example you have seen the taxicab cost modeled in words, an equation, a table and in graphical form. Whenever possible, ensure that you can link these four representations together to continually build your skills. It is important to note that you will not always be able to find all 4 representations for a problem and so being able to work with all 4 forms is very important.
Section 2.1 Linear Functions101
Linear Function A linear function is a function whose graph produces a line. Linear functions can always be written in the form or f ( x) = b + mx f ( x) = mx + b ; they’re equivalent where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change of the function Many people like to write linear functions in the form f ( x) = b + mx because it corresponds to the way we tend to speak: “The output starts at b and increases at a rate of m.” For this reason alone we will use the f ( x) = b + mx form for many of the examples, but remember they are equivalent and can be written correctly both ways. Slope and Increasing/Decreasing m is the constant rate of change of the function (also called slope). The slope determines if the function is an increasing function or a decreasing function. f ( x) = b + mx is an increasing function if m > 0 f ( x) = b + mx is a decreasing function if m < 0 If m = 0 , the rate of change zero, and the function f ( x) =b + 0 x =b is just a horizontal line passing through the point (0, b), neither increasing nor decreasing. Example 1 Marcus currently owns 200 songs in his iTunes collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his iTunes collection as a function of the number of months, m. How many songs will he own in a year? The initial value for this function is 200, since he currently owns 200 songs, so N (0) = 200 . The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. With this information, we can write the formula: N (m) = 200 + 15m . N(m) is an increasing linear function. With this formula we can predict how many songs he will have in 1 year (12 months): N (12) = 200 + 15(12) = 200 + 180 = 380 . Marcus will have 380 songs in 12 months.
102 Chapter 2 Try it Now 1. If you earn $30,000 per year and you spend $29,000 per year write an equation for the amount of money you save after y years, if you start with nothing. “The most important thing, spend less than you earn! 2” Calculating Rate of Change Given two values for the input, x1 and x 2 , and two corresponding values for the output, y1 and y 2 , or a set of points, ( x1 , y1 ) and ( x 2 , y 2 ) , if we wish to find a linear function that contains both points we can calculate the rate of change, m: change in output ∆y y 2 − y1 m= = = change in input ∆x x 2 − x1 Rate of change of a linear function is also called the slope of the line. Note in function notation, y1 = f ( x1 ) and y 2 = f ( x 2 ) , so we could equivalently write f ( x2 ) − f ( x1 ) m= x2 − x1 Example 2 The population of a city increased from 23,400 to 27,800 between 2002 and 2006. Find the rate of change of the population during this time span. The rate of change will relate the change in population to the change in time. The population increased by 27800 − 23400 = 4400 people over the 4 year time interval. To find the rate of change, the number of people per year the population changed by: 4400 people people = 1100 people per year = 1100 4 years year Notice that we knew the population was increasing, so we would expect our value for m to be positive. This is a quick way to check to see if your value is reasonable. Example 3 The pressure, P, in pounds per square inch (PSI) on a diver depends upon their depth below the water surface, d, in feet, following the equation P(d ) = 14.696 + 0.434d . Interpret the components of this function.
2
http://www.thesimpledollar.com/2009/06/19/rule-1-spend-less-than-you-earn/
Section 2.1 Linear Functions103 output pressure PSI . This = = input depth ft tells us the pressure on the diver increases by 0.434 PSI for each foot their depth increases. The rate of change, or slope, 0.434 would have units
The initial value, 14.696, will have the same units as the output, so this tells us that at a depth of 0 feet, the pressure on the diver will be 14.696 PSI. Example 4 If f (x) is a linear function, f (3) = −2 , and f (8) = 1 , find the rate of change. f (3) = −2 tells us that the input 3 corresponds with the output -2, and f (8) = 1 tells us that the input 8 corresponds with the output 1. To find the rate of change, we divide the change in output by the change in input:
m=
change in output 1 − (−2) 3 = = . If desired we could also write this as m = 0.6 change in input 8−3 5
Note that it is not important which pair of values comes first in the subtractions so long as the first output value used corresponds with the first input value used. Try it Now 2. Given the two points (2, 3) and (0, 4), find the rate of change. Is this function increasing or decreasing? We can now find the rate of change given two input-output pairs, and can write an equation for a linear function once we have the rate of change and initial value. If we have two input-output pairs and they do not include the initial value of the function, then we will have to solve for it. Example 5 Write an equation for the linear function graphed to the right. Looking at the graph, we might notice that it passes through the points (0, 7) and (4, 4). From the first value, we know the initial value of the function is b = 7, so in this case we will only need to calculate the rate of change:
104 Chapter 2 m=
4−7 −3 = 4−0 4
This allows us to write the equation: 3 f ( x) = 7 − x 4 Example 6 If f (x) is a linear function, f (3) = −2 , and f (8) = 1 , find an equation for the function. 3 . In this case, we do not 5 know the initial value f (0) , so we will have to solve for it. Using the rate of change, 3 we know the equation will have the form f ( x) = b + x . Since we know the value of 5 the function when x = 3, we can evaluate the function at 3.
In example 3, we computed the rate of change to be m =
3 f (3) = b + (3) 5 3 − 2 = b + (3) 5 9 − 19 b = −2 − = 5 5
Since we know that f (3) = −2 , we can substitute on the left side This leaves us with an equation we can solve for the initial value
Combining this with the value for the rate of change, we can now write a formula for this function: − 19 3 f ( x) = + x 5 5 Example 7 Working as an insurance salesperson, Ilya earns a base salary and a commission on each new policy, so Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies, and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation. The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. 920 − 760 160 m= = = 80 5−3 2
Section 2.1 Linear Functions105 Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy; Ilya earns a commission of $80 for each policy sold during the week. We can then solve for the initial value then when n = 3, I (3) = 760 , giving I (n) = b + 80n this allows us to solve for b 760 = b + 80(3) b = 760 − 80(3) = 520 This value is the starting value for the function. This is Ilya’s income when n = 0, which means no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. Writing the final equation: I (n) = 520 + 80n Our final interpretation is: Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold each week. Flashback Looking at Example 7: Determine the independent and dependent variables. What is a reasonable domain and range? Is this function one-to-one? Try it Now 3. The balance in your college payment account, C, is a function of the number of quarters, q, you attend. Interpret the function C(a) = 20000 – 4000q in words. How many quarters of college can you pay for until this account is empty? Example 8 Given the table below write a linear equation that represents the table values w, number of weeks P(w), number of rats
0
2
4
6
1000
1080
1160
1240
We can see from the table that the initial value of rats is 1000 so in the linear format P( w)= b + mw , b = 1000. Rather than solving for m, we can notice from the table that the population goes up by 80 for every 2 weeks that pass. This rate is consistent from week 0, to week 2, 4, and 6.
106 Chapter 2 The rate of change is 80 rats per 2 weeks. This can be simplified to 40 rats per week and we can write P( w)= b + mw as P( w) = 1000 + 40w If you didn’t notice this from the table you could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240), 1240 − 1080 160 m = = = 40 rats per week 6−2 4 Important Topics of this Section Definition of Modeling Definition of a linear function Structure of a linear function Increasing & Decreasing functions Finding the vertical intercept (0, b) Finding the slope/rate of change, m Interpreting linear functions Try it Now Answers 1. S ( y ) = 30,000 y − 29,000 y = 1000 y $1000 is saved each year. 4−3 1 1 2. m = = = − ; Decreasing because m < 0 0−2 −2 2 3. Your College account starts with $20,000 in it and you withdraw $4,000 each quarter (or your account contains $20,000 and decreases by $4000 each quarter.) You can pay for 5 quarters before the money in this account is gone. Flashback Answers n (number of policies sold) is the independent variable I(n) (weekly income as a function of policies sold) is the dependent variable. A reasonable domain is (0, 15)* A reasonable range is ($540, $1740)* * answers may vary given reasoning is stated; 15 is an arbitrary upper limit based on selling 3 policies per day in a 5 day work week and $1740 corresponds with the domain. Yes this function is one-to-one
Section 2.1 Linear Functions107
Section 2.1 Exercises 1. A town's population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1700 people each year. Write an equation, , P ( t ) for the population t years after 2003. 2. A town's population has been growing linearly. In 2005, the population was 69,000, and the population has been growing by 2500 people each year. Write an equation, , P ( t ) for the population t years after 2005. 3. Sonya is currently 10 miles from home, and is walking further away at 2 miles per hour. Write an equation for her distance from home t hours from now. 4. A boat is 100 miles away from the marina, sailing directly towards it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. 5. Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding n rides? 6. At noon, a barista notices she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift? Determine if each function is increasing or decreasing 8. g ( x= 7. f ( x= ) 4x + 3 ) 5x + 6 9. a ( x )= 5 − 2 x
10. b ( x )= 8 − 3 x
−2 x + 4 11. h ( x ) =
−4 x + 1 12. k ( x ) =
1 x −3 2 1 15. n ( x ) = − x−2 3
14. p ( = x)
13. j ( = x)
1 x −5 4 3 16. m ( x ) = − x+3 8
Find the slope of the line that passes through the two given points 17. (2, 4) and (4, 10) 18. (1, 5) and (4, 11) 19. (-1,4) and (5, 2) 20. (-2, 8) and (4, 6) 21. (6,11) and (-4,3) 22. (9,10) and (-6,-12)
108 Chapter 2 Find the slope of the lines graphed
23.
24.
25. Sonya is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate? 26. A gym membership with two personal training sessions costs $125, while gym membership with 5 personal training sessions costs $260. What is the rate for personal training sessions? 27. A city's population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the slope of the population growth (or decline) and make a statement about the population rate of change in people per year. 28. A city's population in the year 1958 was 2,113,000. In 1991 the population was 2,099,800. Compute the slope of the population growth (or decline) and make a statement about the population rate of change in people per year. 29. A phone company charges for service according to the formula: C (n= ) 24 + 0.1n , where n is the number of minutes talked, and C (n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 30. A phone company charges for service according to the formula: C (n= ) 26 + 0.04n , where n is the number of minutes talked, and C (n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 31. Terry is skiing down a steep hill. Terry's elevation, E (t ) , in feet after t seconds is given by E = (t ) 3000 − 70t . Write a complete sentence describing Terry’s starting elevation and how it is changing over time.
Section 2.1 Linear Functions109 32. Maria is climbing a mountain. Maria's elevation, E (t ) , in feet after t minutes is given by E= ( t ) 1 200 + 40t . Write a complete sentence describing Maria’s starting elevation and how it is changing over time. Given each set of information, find a linear equation satisfying the conditions, if possible 33. f (−5) 4 34. f (−1) 4 = − , and f (5) 2 =, and f (5) 1 = = 35. Passes through (2, 4) and (4, 10)
36. Passes through (1, 5) and (4, 11)
37. Passes through (-1,4) and (5, 2)
38. Passes through (-2, 8) and (4, 6)
39. x intercept at (-2, 0) and y intercept at (0, -3) 40. x intercept at (-5, 0) and y intercept at (0, 4) Find an equation for the function graphed
41.
42.
43.
44.
45. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1000 shirts can be sold at a price of $30 , while 3000 shirts can be sold at a price of $22 . Find a linear equation in the form = p mn + b that gives the price p they can charge for n shirts.
110 Chapter 2 46. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form = y mn + b that gives the yield when n stalks are planted. 47. Which of the following tables could represent a linear function? For each that could be linear, find a linear equation models the data. x 0 5 10 15
g(x) 5 -10 -25 -40
x 0 5 10 15
h(x) 5 30 105 230
x 0 5 10 15
f(x) -5 20 45 70
x 5 10 20 25
k(x) 13 28 58 73
48. Which of the following tables could represent a linear function? For each that could be linear, find a linear equation models the data. x 0 2 4 6
g(x) 6 -19 -44 -69
x 2 4 8 10
h(x) 13 23 43 53
x 2 4 6 8
f(x) -4 16 36 56
x 0 2 6 8
k(x) 6 31 106 231
49. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was -23 Celsius (-23oC). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees, oF, but you could not find a reference with the correct formulas. You then remembered that the relationship between oF and oC is linear. [UW] a. Using this and the knowledge that 32oF = 0 oC and 212 oF = 100 oC, find an equation that computes Celsius temperature in terms of Fahrenheit; i.e. an equation of the form C = “an expression involving only the variable F.” b. Likewise, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F = “an expression involving only the variable C.” c. How cold was it in Oslo in oF?
Section 2.2 Graphs of Linear Functions 111
Section 2.2 Graphs of Linear Functions When we are working with a new function, it is useful to know as much as we can about the function: its graph, where the function is zero, and any other special behaviors of the function. We will begin this exploration of linear functions with a look at graphs. When graphing a linear function, there are three basic ways to graph it: 1) By plotting points (at least 2) and drawing a line through the points 2) Using the initial value (output when x = 0) and rate of change (slope) 3) Using transformations of the identity function f ( x) = x Example 1 Graph f ( x) = 5 −
2 x by plotting points 3
In general, we evaluate the function at two or more inputs to find at least two points on the graph. Usually it is best to pick input values that will “work nicely” in the equation. 2 In this equation, multiples of 3 will work nicely due to the in the equation, and of 3 course using x = 0 to get the vertical intercept. Evaluating f(x) at x = 0, 3 and 6: 2 f ( 0) = 5 − ( 0) = 5 3 2 f (3) = 5 − (3) = 3 3 2 f (6) = 5 − (6) = 1 3 These evaluations tell us that the points (0,5), (3,3), and (6,1) lie on the graph of the line. Plotting these points and drawing a line through them gives us the graph
112 Chapter 2 When using the initial value and rate of change to graph, we need to consider the graphical interpretation of these values. Remember the initial value of the function is the output when the input is zero, so in the equation f ( x) = b + mx , the graph includes the point (0, b). On the graph, this is the vertical intercept – the point where the graph crosses the vertical axis. For the rate of change, it is helpful to recall that we calculated this value as change of output m= change of input From a graph of a line, this tells us that if we divide the vertical difference, or rise, of the function outputs by the horizontal difference, or run, of the inputs, we will obtain the rate of change, also called slope of the line. m=
change of output rise = change of input run
Notice that this ratio is the same regardless of which two points we use rise 2, run 4 m = 2/4 = ½ run 2, rise 1 m=½ run 2, rise 1 m=½
Graphical Interpretation of a Linear Equation Graphically, in the equation f ( x) = b + mx b is the vertical intercept of the graph and tells us we can start our graph at (0, b) m is the slope of the line and tells us how far to rise & run to get to the next point Once we have at least 2 points, we can extend the graph of the line to the left and right. Example 2 Graph f ( x) = 5 −
2 x using the vertical intercept and slope. 3
The vertical intercept of the function is (0, 5), giving us a point on the graph of the line.
Section 2.2 Graphs of Linear Functions 113 2 . This tells us that for every 3 units the graph “runs” in the horizontal, 3 the vertical “rise” decreases by 2 units. In graphing, we can use this by first plotting our vertical intercept on the graph, then using the slope to find a second point. From the initial value (0, 5) the slope tells us that if we move to the right 3, we will move down 2, moving us to the point (3, 3). We can continue this again to find a third point at (6, 1). Finally, extend the line to the left and right, containing these points.
The slope is −
Try it Now 1. Consider that the slope -2/3 could also be written as point on the graph that has a negative x value.
2 2 . Using , find another −3 −3
Another option for graphing is to use transformations of the identity function f ( x) = x . In the equation f ( x) = mx , the m is acting as the vertical stretch of the identity function. When m is negative, there is also a vertical reflection of the graph. Looking at some examples: f ( x) = 3x f ( x) = 2 x f ( x) = x
f ( x) =
1 x 2
1 f ( x) = x 3
1 f ( x) = − x 2
f ( x) = −2 x
f ( x) = − x
114 Chapter 2 In f ( x) = mx + b , the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Some examples: f ( x)= x + 4
f ( x)= x + 2 f ( x) = x
f ( x)= x − 2 f ( x)= x − 4
Using Vertical Stretches or Compressions along with Vertical Shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way for you to graph this type of function, make sure you practice each method. Example 3 Graph f ( x) = −3 +
1 x using transformations. 2
The equation is the graph of the identity function vertically compressed by ½ and vertically shifted down 3. Vertical compression
combined with Vertical shift
Section 2.2 Graphs of Linear Functions 115 Notice how this nicely compares to the other method where the vertical intercept is found at (0, -3) and to get to another point we rise (go up vertically) by 1 unit and run (go horizontally) by 2 units to get to the next point (2,-2), and the next one (4, -1). In these three points (0,-3), (2, -2), and (4, -1), the output values change by +1, and the x values change by +2, corresponding with the slope m = 1/2. Example 4 Match each equation with one of the lines in the graph below f ( x) = 2 x + 3 g ( x) = 2 x − 3 h( x) = −2 x + 3 j ( x) =
1 x+3 2
Only one graph has a vertical intercept of -3, so we can immediately match that graph with g(x). For the three graphs with a vertical intercept at 3, only one has a negative slope, so we can match that line with h(x). Of the other two, the steeper line would have a larger slope, so we can match that graph with equation f(x), and the flatter line with the equation j(x).
j (= x)
1 x+3 2
f ( x= ) 2x + 3
g ( x= ) 2x − 3
h( x ) = −2 x + 3
116 Chapter 2 In addition to understanding the basic behavior of a linear function (increasing or decreasing, recognizing the slope and vertical intercept), it is often helpful to know the horizontal intercept of the function – where it crosses the horizontal axis. Finding Horizontal Intercept The horizontal intercept of the function is where the graph crosses the horizontal axis. If a function has a horizontal intercept, you can always find it by solving f(x) = 0. Example 5 Find the horizontal intercept of f ( x) = −3 +
1 x 2
Setting the function equal to zero to find what input will put us on the horizontal axis, 1 0 = −3 + x 2 1 3= x 2 x=6 The graph crosses the horizontal axis at (6,0) There are two special cases of lines: a horizontal line and a vertical line. In a horizontal line like the one graphed to the right, notice that between any two points, the change in the outputs is 0. In the slope equation, the numerator will be 0, resulting in a slope of 0. Using a slope of 0 in the f ( x) = b + mx , the equation simplifies to f ( x) = b . Notice a horizontal line has a vertical intercept, but no horizontal intercept (unless it’s the line f(x) = 0). In the case of a vertical line, notice that between any two points, the change in the inputs is zero. In the slope equation, the denominator will be zero, and you may recall that we cannot divide by the zero; the slope of a vertical line is undefined. You might also notice that a vertical line is not a function. To write the equation of vertical line, we simply write input=value, like x = b . Notice a vertical line has a horizontal intercept, but no vertical intercept (unless it’s the line x = 0).
Section 2.2 Graphs of Linear Functions 117 Horizontal and Vertical Lines Horizontal lines have equations of the form f ( x) = b Vertical lines have equations of the form x = a Example 6 Write an equation for the horizontal line graphed above. This line would have equation f ( x) = 2 Example 7 Write an equation for the vertical line graphed above. This line would have equation x = 2 Try it Now 2. Describe the function f ( x) = 6 − 3 x in terms of transformations of the identity function and find its horizontal intercept. Parallel and Perpendicular Lines When two lines are graphed together, the lines will be parallel if they are increasing at the same rate – if the rates of change are the same. In this case, the graphs will never cross (unless they’re the same line). Parallel Lines Two lines are parallel if the slopes are equal (or, if both lines are vertical). In other words, given two linear equations f ( x) = b + m1 x and g ( x) = b + m2 x The lines will be parallel if m1 = m2 Example 8 Find a line parallel to f ( x) = 6 + 3 x that passes through the point (3, 0) We know the line we’re looking for will have the same slope as the given line, m = 3. Using this and the given point, we can solve for the new line’s vertical intercept: then at (3, 0), g ( x) = b + 3x 0 = b + 3(3) b = −9 The line we’re looking for is g ( x) = −9 + 3 x
118 Chapter 2 If two lines are not parallel, one other interesting possibility is that the lines are perpendicular, which means the lines form a right angle (90 degree angle – a square corner) where they meet. In this case, the slopes when multiplied together will equal -1. Solving for one slope leads us to the definition: Perpendicular Lines Given two linear equations f ( x) = b + m1 x and g ( x) = b + m2 x −1 The lines will be perpendicular if m1 m2 = −1 , and so m2 = m1 We often say the slope of a perpendicular line is the “negative reciprocal” of the other line’s slope. Example 9 Find the slope of a line perpendicular to a line with: a) a slope of 2.
b) a slope of -4.
c) a slope of
2 . 3
−1 2 −1 1 If the original line had slope -4, the perpendicular line’s slope would be m2 = = −4 4 2 −1 − 3 If the original line had slope , the perpendicular line’s slope would be m2 = = 2 3 2 3
If the original line had slope 2, the perpendicular line’s slope would be m2 =
Example 10 Find the equation of a line perpendicular to f ( x) = 6 + 3 x and passing through the point (3, 0) The original line has slope m = 3. The perpendicular line will have slope m = Using this and the given point, we can find the equation for the line. 1 then at (3, 0), g ( x) = b − x 3 1 0 = b − (3) 3 b =1 1 The line we’re looking for is g ( x) = 1 − x 3
−1 . 3
Section 2.2 Graphs of Linear Functions 119 Try it Now 3. Given the line h(t ) = −4 + 2t , find an equation for the line passing through (0, 0) that is: a) parallel to h(t). b) perpendicular to h(t). Example 12 A line passes through the points (-2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5). From the two given points on the reference line, we can calculate the slope of that line: 5−6 −1 m1 = = 4 − (−2) 6 The perpendicular line will have slope −1 m2 = =6 −1 6 We can then solve for the vertical intercept that makes the line pass through the desired point: then at (4, 5), g ( x) = b + 6 x 5 = b + 6( 4) b = −19 Giving the line g ( x) = −19 + 6 x Intersections of Lines The graphs of two lines will intersect if they are not parallel. They will intersect at the point that satisfies both equations. To find this point when the equations are given as functions, we can solve for an input value so that f ( x) = g ( x) . In other words, we can set the formulas for the lines equal, and solve for the input that satisfies the equation. Example 13 Find the intersection of the lines h(t ) = 3t − 4 and j (t ) = 5 − t Setting h(t ) = j (t ) , 3t − 4 = 5 − t 4t = 9 9 t= 4 This tells us the lines intersect when the input is
9 . 4
120 Chapter 2 We can then find the output value of the intersection point by evaluating either function at this input 9 11 9 j = 5 − = 4 4 4 9 11 These lines intersect at the point , . Looking at the graph, this result seems 4 4 reasonable. h(t)
j(t)
Two parallel lines can also intersect if they happen to be the same line. In that case, they intersect at every point on the lines. Try it Now 4. Look at the graph in example 13 above and answer the following for the function j(t): a. Vertical intercept coordinates b. Horizontal intercepts coordinates c. Slope d. Is j(t) parallel or perpendicular to h(t) (or neither) e. Is j(t) an Increasing or Decreasing function (or neither) f. Write a transformation description from the identity toolkit function f(x) = x Finding the intersection allows us to answer other questions as well, such as discovering when one function is larger than another. Example 14 Using the functions from the previous example, for what values of t is h(t ) > j (t ) To answer this question, it is helpful first to know where the functions are equal, since that is the point where h(t) could switch from being greater to smaller than j(t) or vice9 versa. From the previous example, we know the functions are equal at t = . 4
Section 2.2 Graphs of Linear Functions 121 By examining the graph, we can see that h(t), the function with positive slope, is going to be larger than the other function to the right of the intersection. So h(t ) > j (t ) when 9 t> 4 Important Topics of this Section Methods for graphing linear functions Another name for slope = rise/run Horizontal intercepts (a,0) Horizontal lines Vertical lines Parallel lines Perpendicular lines Intersecting lines Try it Now Answers 1. (-3,7) found by starting at the vertical intercept, going up 2 units and 3 in the negative horizontal direction. You could have also answered, (-6, 9) or (-9, 11) etc… 2. Vertically stretched by a factor of 3, Vertically flipped (flipped over the x axis), Vertically shifted up by 6 units. 6-3x=0 when x=2 1 3. Parallel f (t ) = 2t ; Perpendicular g (t ) = − t 2 4. Given j(t) = 5-t a. (0,5) b. (5,0) c. Slope -1 d. Neither parallel nor perpendicular e. Decreasing function f. Given the identity function, perform a vertical flip (over the t axis) and shift up 5 units.
122 Chapter 2
Section 2.2 Exercises Match each linear equation with its graph 1.
f ( x ) =− x − 1
2.
f ( x) = −2 x − 1
4.
1 f ( x) = − x −1 2 f ( x) = 2
5.
f ( x )= 2 + x
6.
f ( x= ) 3x + 2
3.
A
C
D F
Sketch a line with the given features 7. An x-intercept of (-4, 0) and y-intercept of (0, -2) 8. An x-intercept of (-2, 0) and y-intercept of (0, 4) 9. A vertical intercept of (0, 7) and slope − 10. A vertical intercept of (0, 3) and slope
3 2
2 5
11. Passing through the points (-6,-2) and (6,-6) 12. Passing through the points (-3,-4) and (3,0) Sketch the graph of each equation 13. f ( x ) = −2 x − 1 15. h ( = x)
1 x+2 3
B
14. g ( x ) = −3 x + 2 16. k ( = x)
2 x −3 3
17. k ( t )= 3 + 2t
18. p ( t ) =−2 + 3t
19. x = 3
20. x = −2
21. r ( x ) = 4
22. q ( x ) = 3
E
Section 2.2 Graphs of Linear Functions 123 23. If g ( x) is the transformation of f ( x) = x after a vertical compression by 3 / 4 , a shift left by 2, and a shift down by 4 a. Write an equation for g ( x ) b. What is the slope of this line? c. Find the vertical intercept of this line. 24. If g ( x) is the transformation of f ( x) = x after a vertical compression by 1/ 3 , a shift right by 1, and a shift up by 3 a. Write an equation for g ( x ) b. What is the slope of this line? c. Find the vertical intercept of this line. Write the equation of the line shown
25.
26.
27.
28.
Find the horizontal and vertical intercepts of each equation 30. g ( x= 29. f ( x ) =− x + 2 ) 2x + 4 31. h ( x= ) 3x − 5
32. k ( x ) = −5 x + 1
33. −2 x + 5 y = 20
34. 7 x + 2 y = 56
124 Chapter 2 Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular or neither? 35. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1,19) and (8, −71) 36. Line 1: Passes through (−8, 55) and (10, 89) − Line 2: Passes through (9, 44) and (4, −14) − 37. Line 1: Passes through (2,3) and (4, −1) Line 2: Passes through (6,3) and (8,5) 38. Line 1: Passes through (1, 7) and (5,5) Line 2: Passes through (−1, −3) and (1,1) 39. Line 1: Passes through (0, 5) and (3,3) Line 2: Passes through (1, −5) and (3, −2) 40. Line 1: Passes through (2,5) and (5, −1) Line 2: Passes through (−3, 7) and (3, −5) −5 x − 3 and passing through the point 41. Write an equation for a line parallel to f ( x ) = (2,-12) 42. Write an equation for a line parallel to g ( x= ) 3 x − 1 and passing through the point (4,9) 43. Write an equation for a line perpendicular to h(t ) = −2t + 4 and passing through the point (-4,-1) 44. Write an equation for a line perpendicular to p (t= ) 3t + 4 and passing through the point (3,1) 45. Find the point at which the line f ( x) = −2 x − 1 intersects the line g ( x) = − x 46. Find the point at which the line f ( x= ) 2 x + 5 intersects the line g ( x) = −3 x − 5
Section 2.2 Graphs of Linear Functions 125 4 274 47. Use algebra to find the point at which the line f ( x ) = intersects the line − x+ 5 25 9 73 h(= x) x + 4 10
48. Use algebra to find the point at which the line f ( = x) g (= x)
4 31 x + 3 5
7 457 intersects the line x + 4 60
49. A car rental company offers two plans for renting a car. Plan A: 30 dollars per day and 18 cents per mile Plan B: 50 dollars per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 50. A cell phone company offers two data options for its prepaid phones Pay per use: $0.002 per Kilobyte (KB) used Data Package: $5 for 5 Megabytes (5120 Kilobytes) + $0.002 per addition KB Assuming you will use less than 5 Megabytes, under what circumstances will the data package save you money? 51. Sketch an accurate picture of the line having equation f ( x )= 2 −
1 x . Let c be an 2
unknown constant. [UW] a. Find the point of intersection between the line you have graphed and the line g ( x ) = 1 + cx ; your answer will be a point in the xy plane whose coordinates involve the unknown c. b. Find c so that the intersection point in (a) has x-coordinate 10. c. Find c so that the intersection point in (a) lies on the x-axis.
126 Chapter 2
Section 2.3 Modeling with Linear Functions When modeling scenarios with a linear function and solving problems involving quantities changing linearly, we typically follow the same problem solving strategies that we would use for any type of function: Problem solving strategy 1) Identify changing quantities, and then carefully and clearly define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2) Carefully read the problem to identify important information. Look for information giving values for the variables, or values for parts of the functional model, like slope and initial value. 3) Carefully read the problem to identify what we are trying to find, identify, solve, or interpret. 4) Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table or even finding a formula for the function being used to model the problem. 5) When needed, find a formula for the function. 6) Solve or evaluate using the formula you found for the desired quantities. 7) Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8) Clearly convey your result using appropriate units, and answer in full sentences when appropriate. Example 1 Emily saved up $3500 for her summer visit to Seattle. She anticipates spending $400 each week on rent, food, and fun. Find and interpret the horizontal intercept and determine a reasonable domain and range for this function. In the problem, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can define our variables, including units. Output: M, money remaining, in dollars Input: t, time, in weeks Reading the problem, we identify two important values. The first, $3500, is the initial value for M. The other value appears to be a rate of change – the units of dollars per week match the units of our output variable divided by our input variable. She is spending money each week, so you should recognize that the amount of money remaining is decreasing each week and the slope is negative.
Section 2.3 Modeling with Linear Functions 127 To answer the first question, looking for the horizontal intercept, it would be helpful to have an equation modeling this scenario. Using the intercept and slope provided in the problem, we can write the equation: M (t ) = 3500 − 400t . To find the horizontal intercept, we set the output to zero, and solve for the input: 0 = 3500 − 400t 3500 t= = 8.75 400 The horizontal intercept is 8.75 weeks. Since this represents the input value where the output will be zero, interpreting this, we could say: Emily will have no money left after 8.75 weeks. When modeling any real life scenario with functions, there is typically a limited domain over which that model will be valid – almost no trend continues indefinitely. In this case, it certainly doesn’t make sense to talk about input values less than zero. It is also likely that this model is not valid after the horizontal intercept (unless Emily’s going to start using a credit card and go into debt). The domain represents the set of input values and so the reasonable domain for this function is 0 ≤ t ≤ 8.75 . However, in a real world scenario, the rental might be weekly or nightly. She may not be able to stay a partial week and so all options should be considered. Emily could stay in Seattle for 0 to 8 full weeks (and a couple of days), but would have to go into debt to stay 9 full weeks, so restricted to whole weeks, a reasonable domain without going in to debt would be 0 ≤ t ≤ 8 , or 0 ≤ t ≤ 9 if she went into debt to finish out the last week. The range represents the set of output values and she starts with $3500 and ends with $0 after 8.75 weeks so the corresponding range is 0 ≤ M (t ) ≤ 3500 . If we limit the rental to whole weeks however, if she left after 8 weeks because she didn’t have enough to stay for a full 9 weeks, she would have M(8) = 3500 -400(8) = $300 dollars left after 8 weeks, giving a range of 300 ≤ M (t ) ≤ 3500 . If she wanted to stay the full 9 weeks she would be $100 in debt giving a range of − 100 ≤ M (t ) ≤ 3500 . Most importantly remember that domain and range are tied together, and what ever you decide is most appropriate for the domain (the independent variable) will dictate the requirements for the range (the dependent variable).
128 Chapter 2 Example 2 Jamal is choosing between two moving companies. The first, U-Haul, charges an upfront fee of $20, then 59 cents a mile. The second, Budget, charges an up-front fee of $16, then 63 cents a mile 3. When will U-Haul be the better choice for Jamal? The two important quantities in this problem are the cost, and the number of miles that are driven. Since we have two companies to consider, we will define two functions: Input: m, miles driven Outputs: Y(m): cost, in dollars, for renting from U-Haul B(m): cost, in dollars, for renting from Budget Reading the problem carefully, it appears that we were given an initial cost and a rate of change for each company. Since our outputs are measured in dollars but the costs per mile given in the problem are in cents, we will need to convert these quantities to match our desired units: $0.59 a mile for U-Haul, and $0.63 a mile for Budget. Looking to what we’re trying to find, we want to know when U-Haul will be the better choice. Since all we have to make that decision from is the costs, we are looking for when U-Haul will cost less, or when Y (m) < B(m) . The solution pathway will lead us to find the equations for the two functions, find the intersection, then look to see where the Y(m) function is smaller. Using the rates of change and initial charges, we can write the equations: Y (m) = 20 + 0.59m B(m) = 16 + 0.63m These graphs are sketched to the right, with Y(m) drawn dashed. To find the intersection, we set the equations equal and solve: Y(m) = B(m) 20 + 0.59m = 16 + 0.63m 4 = 0.04m m = 100 This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that Y(m) is growing at a slower rate, we can conclude that U-Haul will be the cheaper price when more than 100 miles are driven.
3
Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/
Section 2.3 Modeling with Linear Functions 129 Example 3 A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. If this trend continues, a. Predict the population in 2013 b. When will the population reach 15000? The two changing quantities are the population and time. While we could use the actual year value as the input quantity, doing so tends to lead to very ugly equations, since the vertical intercept would correspond to the year 0, more than 2000 years ago! To make things a little nicer, and to make our lives easier too, we will define our input as years since 2004: Input: t, years since 2004 Output: P(t), the town’s population The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have “given” ourselves the vertical intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100). To predict the population in 2013 (t = 9), we would need an equation for the population. Likewise, to find when the population would reach 15000, we would need to solve for the input that would provide an output of 15000. Either way, we need an equation. To find it, we start by calculating the rate of change: 8100 − 6200 1900 m= = = 380 people per year 5−0 5 Since we already know the vertical intercept of the line, we can immediately write the equation: P(t ) = 6200 + 380t To predict the population in 2013, we evaluate our function at t = 9 P(9) = 6200 + 380(9) = 9620 If the trend continues, our model predicts a population of 9,620 in 2013. To find when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 6200 + 380t 8800 = 380t t ≈ 23.158 Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.
130 Chapter 2 Example 4 Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a twoway radio with a range of 2 miles. How long after they start walking will they fall out of radio contact? In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: how long will it take them to be 2 miles apart? In this problem, our changing quantities are time and the two peoples’ positions, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define Input: t, time in hours. Since it is not obvious how to define our output variables, we’ll start by drawing a picture. Anna walking east, 4 miles/hour
Distance between them Emanuel walking south, 3 miles/hour Because of the complexity of this question, it may be helpful to introduce some intermediary variables. These are quantities that we aren’t directly interested in, but seem important to the problem. For this problem, Anna’s and Emanuel’s distances from the starting point seem important. To notate these, we are going to define a coordinate system, putting the “starting point” at the intersection where they both started, then we’re going to introduce a variable, A, to represent Anna’s position, and define it to be a measurement from the starting point, in the eastward direction. Likewise, we’ll introduce a variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system we specified both the origin, or starting point, of the measurement, as well as the direction of measure. While we’re at it, we’ll define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the picture is often helpful: Looking at the variables on the picture, we remember we need to know how long it takes for D, the distance between them, to equal 2 miles.
Section 2.3 Modeling with Linear Functions 131
A E
D c
a Seeing this picture we remember that in order to find the distance between the two, we can use the Pythagorean Theorem, a property of right triangles.
b a +b = c2 2
2
From here, we can now look back at the problem for relevant information. Anna is walking 4 miles per hour, and Emanuel is walking 3 miles per hour, which are rates of change. Using those, we can write formulas for the distance each has walked. They both start at the same intersection and so when t = 0, the distance travelled by each person should also be 0, so given the rate for each, and the initial value for each, we get: A(t ) = 4t E (t ) = 3t
Using the Pythagorean theorem we get: D= (t ) 2 A(t ) 2 + E (t ) 2 D(t ) 2 = (4t ) 2 + (3t ) 2 = 16t 2 + 9t 2 = 25t 2
substitute in the function formulas solve for D(t) using the square root
D(t ) = ± 25t 2 = ±5 t Since in this scenario we are only considering positive values of t and our distance D(t) will always be positive, we can simplify this answer to D(t ) = 5t Interestingly, the distance between them is also a linear function. Using it, we can now answer the question of when the distance between them will reach 2 miles: D(t ) = 2 5t = 2 2 t= = 0.4 5 They will fall out of radio contact in 0.4 hours, or 24 minutes.
132 Chapter 2 Example 5 There is currently a straight road leading from the town of Westborough to a town 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? It might help here to draw a picture of the situation. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts the other town at coordinates (30, 10), and Eastborough at (20, 0).
Other town (30, 10) (0, 0) Westborough
20 miles
(20, 0) Eastborough
Using this point along with the origin, we can find the slope of the line from 10 − 0 1 Westborough to the other town: m = = . This gives the equation of the road 30 − 0 3 1 from Westborough to the other town to be W ( x) = x . 3 From this, we can determine the perpendicular road to Eastborough will have slope m = −3 . Since the town of Eastborough is at the point (20, 0), we can find the equation: plug in the point (20, 0) E ( x) = −3 x + b 0 = −3(20) + b b = 60 E ( x) = −3 x + 60 We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, 1 x = −3 x + 60 3 10 x = 60 3 10 x = 180 Substituting this back into W(x) x = 18 1 y = W (18) = (18) = 6 3 The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction: dist = (18 − 0) 2 + (6 − 0) 2 ≈ 18.934 miles.
Section 2.3 Modeling with Linear Functions 133 Important Topics of this Section The problem solving process 1) Identify changing quantities, and then carefully and clearly define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2) Carefully read the problem to identify important information. Look for information giving values for the variables, or values for parts of the functional model, like slope and initial value. 3) Carefully read the problem to identify what we are trying to find, identify, solve, or interpret. 4) Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table or even finding a formula for the function being used to model the problem. 5) When needed, find a formula for the function. 6) Solve or evaluate using the formula you found for the desired quantities. 7) Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8) Clearly convey your result using appropriate units, and answer in full sentences when appropriate.
134 Chapter 2
Section 2.3 Exercises 1. In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2008? b. How long did it take the population to grow from 1001 students to 1697 students? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P, of the school t years after 2000. f. Using your equation, predict the population of the school in 2011. 2. In 2003, a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly. a. How much did the population grow between the year 2003 and 2007? b. How long did it take the population to grow from 1431 people to 2134? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P, of the town t years after 2000. f. Using your equation, predict the population of the town in 2014. 3. A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118. a. Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used. b. Interpret the slope and vertical intercept of the equation. c. Use your equation to find the total monthly cost if 687 minutes are used. 4. A phone company has a monthly cellular data plan where a customer pays a flat monthly fee and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80. a. Find a linear equation for the monthly cost of the data plan as a function of x, the number of MB used. b. Interpret the slope and vertical intercept of the equation. c. Use your equation to find the total monthly cost if 250 MB are used.
Section 2.3 Modeling with Linear Functions 135 5. In 1991, the moose population in a park was measured to be 4360. By 1999, the population was measured again to be 5880. If the population continues to change linearly, a. Find a formula for the moose population, P. b. What does your model predict the moose population to be in 2003? 6. In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. If the population continues to change linearly, a. Find a formula for the owl population, P. b. What does your model predict the owl population to be in 2012? 7. The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010, and is being depleted by about 2.1 billion cubic feet each year. a. Give a linear equation for the remaining federal helium reserves, R, in terms of t, the number of years since 2010. b. In 2015, what will the helium reserves be? c. If the rate of depletion doesn’t change, when will the Federal Helium Reserve be depleted? 8. Suppose the world's current oil reserves are 1820 billion barrels. If, on average, the total reserves is decreasing by 25 billion barrels of oil each year: a. Give a linear equation for the remaining oil reserves, R, in terms of t, the number of years since now. b. Seven years from now, what will the oil reserves be? c. If the rate of depletion isn’t change, when will the world’s oil reserves be depleted? 9. You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable? 10. You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable? 11. When hired at a new job selling jewelry, you are given two pay options: Option A: Base salary of $17,000 a year, with a commission of 12% of your sales Option B: Base salary of $20,000 a year, with a commission of 5% of your sales How much jewelry would you need to sell for option A to produce a larger income?
136 Chapter 2 12. When hired at a new job selling electronics, you are given two pay options: Option A: Base salary of $14,000 a year, with a commission of 10% of your sales Option B: Base salary of $19,000 a year, with a commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income? 13. Find the area of a triangle bounded by the y axis, the line f ( x )= 9 − perpendicular to f ( x) that passes through the origin.
6 x , and the line 7
1 14. Find the area of a triangle bounded by the x axis, the line f ( x= ) 12 − x , and the 3 line perpendicular to f ( x) that passes through the origin.
15. Find the area of a parallelogram bounded by the y axis, the line x = 3 , the line f ( x ) = 1 + 2 x , and the line parallel to f ( x) passing through (2, 7) 16. Find the area of a parallelogram bounded by the x axis, the line g ( x) = 2 , the line f ( x ) = 3 x , and the line parallel to f ( x) passing through (6, 1) 17. If b > 0 and m < 0 , then the line f ( x )= b + mx cuts off a triangle from the first quadrant. Express the area of that triangle in terms of m and b. [UW] 18. Find the value of m so the lines f ( x= ) mx + 5 and g ( x ) = x and the y-axis form a triangle with an area of 10. [UW] 19. The median home values in Mississippi and Hawaii (adjusted for inflation) are shown below. If we assume that the house values are changing linearly, Year Mississippi Hawaii 1950 25200 74400 2000 71400 272700 a. In which state have home values increased at a higher rate? b. If these trends were to continue, what would be the median home value in Mississippi in 2010? c. If we assume the linear trend existed before 1950 and continues after 2000, the two states' median house values will be (or were) equal in what year? (The answer might be absurd)
Section 2.3 Modeling with Linear Functions 137 20. The median home value ins Indiana and Alabama (adjusted for inflation) are shown below. If we assume that the house values are changing linearly, Year Indiana Alabama 1950 37700 27100 2000 94300 85100 a. In which state have home values increased at a higher rate? b. If these trends were to continue, what would be the median home value in Indiana in 2010? c. If we assume the linear trend existed before 1950 and continues after 2000, the two states' median house values will be (or were) equal in what year? (The answer might be absurd) 21. Pam is taking a train from the town of Rome to the town of Florence. Rome is located 30 miles due West of the town of Paris. Florence is 25 miles East, and 45 miles North of Rome. On her trip, how close does Pam get to Paris? [UW] 22. You’re flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 226 km south and 230 km east. Mt. Rainier is located approximately 56 km east and 40 km south of JBLM. If you are flying at a constant speed of 800 km/hr, how long after you depart JBLM will you be the closest to Mt. Rainier?
138 Chapter 2
Section 2.4 Fitting Linear Models to Data In the real world, rarely do things follow trends perfectly. When we expect the trend to behave linearly, or when inspection suggests the trend is behaving linearly, it is often desirable to find an equation to approximate the data. Finding an equation to approximate the data helps us understand the behavior of the data and allows us to use the linear model to make predictions about the data, inside and outside of the data range. Example 1 The table below shows the number of cricket chirps in 15 seconds, and the air temperature, in degrees Fahrenheit 4. Plot this data, and determine whether the data appears to be linearly related. chirps 44 35 20.4 33 31 35 18.5 37 26 Temp 80.5 70.5 57 66 68 72 52 73.5 53 Plotting this data, it appears there may be a trend, and that the trend appears roughly linear, though certainly not perfectly so. Temperature (degrees F)
90 80 70 60 50 40 10
20
30
40
50
Cricket Chirps in 15 seconds
The simplest way to find an equation to approximate this data is to try to “eyeball” a line that seems to fit the data pretty well, then find an equation for that line based on the slope and intercept. You can see from the trend in the data that the number of chirps increases as the temperature increases. As you consider a function for this data you should know that you are looking at an increasing function or a function with a positive slope.
4
Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010
Section 2.4 Fitting Linear Models to Data 139 Flashback 1. a. What descriptive variables would you choose to represent Temperature & Chirps? b. Which variable is the independent variable and which is the dependent variable? c. Based on this data and the graph, what is a reasonable domain & range? d. Based on the data alone, is this function one-to-one, explain? Example 2 Using the table of values from the previous example, find a linear function that fits the data by “eyeballing” a line that seems to fit. On a graph, we could try sketching in a line. The scale on the axes has been adjusted to including the vertical axis in the graph. Using the starting and ending points of our “hand drawn” line, points (0, 30) and (50, 90), 60 this graph has a slope of = m = 1.2 and a 50 vertical intercept at 30, giving an equation of T (c= ) 30 + 1.2c where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit.
90 80 70 60 50 40 30 0
10
20
30
40
This linear equation can then be used to approximate the solution to various questions we might ask about the trend. While the data does not perfectly fall on the linear equation, the equation is our best guess as to how the relationship will behave outside of the values we have data for. There is a difference, though, between making predictions inside the domain and range of values we have data for, and outside that domain and range. Interpolation and Extrapolation Interpolation: When we predict a value inside the domain and range of the data Extrapolation: When we predict a value outside the domain and range of the data For the Temperature as a function of chirps in our hand drawn model above: Interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.
50
140 Chapter 2 Example 3 a) Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss if it is reasonable. b) Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss if it is reasonable. With our cricket data, our number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: T (30) = 30 + 1.2(30) = 66 degrees. Based on the data we have, this value seems reasonable. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation since 40 is outside the range of our data. Using our model: 40 = 30 + 1.2c 10 = 1.2c c ≈ 8.33 Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees. When our model no longer applies after some point, it is sometimes called model breakdown. Try it Now What temperature would you predict if you counted 20 chirps in 15 seconds? Fitting Lines with Technology While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values 5. This technique is called least-square regression, and can be computed by many graphing calculators, spreadsheet software like Excel or Google Docs, statistical software, and many web-based calculators 6.
5
Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. 6 For example, http://www.shodor.org/unchem/math/lls/leastsq.html
Section 2.4 Fitting Linear Models to Data 141 Example 4 Find the least-squares regression line using the cricket chirp data from above. Using the cricket chirp data from earlier, with technology we obtain the equation: = T (c) 30.281 + 1.143c
90 80 70 60
50 Notice that this line is quite similar to the equation we “eyeballed”, but should fit 40 the data better. Notice also that using 30 this equation would change our 0 prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to: T (30) = 30.281 + 1.143(30) = 64.571 ≈ 64.6 degrees.
10
20
30
40
50
Most calculators and computer software will also provide you with the correlation coefficient, a measure of how closely the line fits the data. Correlation Coefficient The correlation coefficient is a value, r, between -1 and 1. r > 0 suggests a positive (increasing) relationship r < 0 suggests a negative (decreasing) relationship The closer the value is to 0, the more scattered the data The closer the value is to 1 or -1, the less scattered the data is The correlation coefficient provides an easy way to get some idea of how close to a line the data falls. We should only compute the correlation coefficient for data that follows a linear pattern; if the data exhibits a non-linear pattern, the correlation coefficient is meaningless. To get a sense for the relationship between the value of r and the graph of the data, here are some large data sets with their correlation coefficients:
142 Chapter 2 Examples of Correlation Coefficient Values
7
Example 5 Calculate the correlation coefficient for our cricket data. Because the data appears to follow a linear pattern, we can use technology to calculate r = 0.9509. Since this value is very close to 1, it suggests a strong increasing linear relationship. Example 6 Gasoline consumption in the US has been increasing steadily. Consumption data from 1994 to 2004 is shown below. 8 Determine if the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.
To make things simpler, a new input variable is introduced, t, representing years since 1994. Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.
Gas Consumption (billions of gallons)
Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136 150 140 130 120 110 100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Years after 1994 http://en.wikipedia.org/wiki/File:Correlation_examples.png 8 http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html 7
Section 2.4 Fitting Linear Models to Data 143 The least-squares regression equation is: C (t ) = 113.318 + 2.209t . Using this to predict consumption in 2008 (t = 14), C (14) = 113.318 + 2.209(14) = 144.244 billions of gallons The model predicts 144.244 billion gallons of gasoline will be consumed in 2008. Try it Now 2. Use the model created by technology in example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation? Important Topics of this Section Fitting linear models to data by hand Fitting linear models to data using technology Interpolation Extrapolation Correlation coefficient Flashback Answers 1. a. T = Temperature, C = Chirps (answers may vary) b. Independent (Chirps) , Dependent (Temperature) c. Reasonable Domain (18.5, 44) , Reasonable Range (52, 80.5) (answers may vary) d. NO, it is not one-to-one, there are two different output values for 35 chirps. Try it Now Answers 1. 54 degrees Fahrenheit 2. 150.871 billion gallons, extrapolation
144 Chapter 2
Section 2.4 Exercises 1. The following is data for the first and second quiz scores for 8 students in a class. Plot the points, then sketch a line that fits the data. 11 10
First Quiz Second Quiz
20 16
24 23
25 28
33 30
42 39
46 40
49 49
2. Eight students were asked to estimate their score on a 10 point quiz. Their estimated and actual scores are given. Plot the points, then sketch a line that fits the data. Predicted Actual
5 6
7 6
6 7
8 8
10 9
9 9
10 10
7 6
Based on each set of data given, calculate the regression line using your calculator or other technology tool, and determine the correlation coefficient. 3.
x 5 7 10 12 15
y 4 12 17 22 24
4.
x y 8 23 15 41 26 53 31 72 56 103
5.
x 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
y 21.9 22.22 22.74 22.26 20.78 17.6 16.52 18.54 15.76 13.68 14.1 14.02 11.94 12.76 11.28 9.1
6.
x 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
y 44.8 43.1 38.8 39 38 32.7 30.1 29.3 27 25.8 24.7 22 20.1 19.8 16.8
Section 2.4 Fitting Linear Models to Data 145 7. A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y). The results of the regression are given below. Use this to predict the number of situps a person who watches 11 hours of TV can do. y=ax+b a=-1.341 b=32.234 r2=0.803 r=-0.896
8. A regression was run to determine if there is a relationship between the diameter of a tree (x, in inches) and the tree’s age (y, in years). The results of the regression are given below. Use this to predict the age of a tree with diameter 10 inches. y=ax+b a=6.301 b=-1.044 r2=0.940 r=-0.970
Match each scatterplot shown below with one of the four specified correlations. 9. r = 0.95 10. r = -0.89 11. r = 0.26 12. r = -0.39
B
A
C
D
13. The US census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given below. Determine if the trend appears linear. If so and the trend continues, in what year will the percentage exceed 35%? Year 1990 1992 1994 1996 1998 2000 2002 2004 2006 2008 Percent 21.3 21.4 22.2 23.6 24.4 25.6 26.7 27.7 28 29.4 Graduates 14. The US import of wine (in hectoliters) for several years is given below. Determine if the trend appears linear. If so and the trend continues, in what year will imports exceed 12,000 hectoliters? Year Imports
1992 1994 1996 1998 2000 2002 2004 2006 2008 2009 2665 2688 3565 4129 4584 5655 6549 7950 8487 9462
146 Chapter 2
Section 2.5 Absolute Value Functions So far in this chapter we have been studying the behavior of linear functions. The Absolute Value Function is a piecewise-defined function made up of two linear functions. The name, Absolute Value Function, should be familiar to you from Section 1.2. In its basic form f ( x) = x it is one of our toolkit functions. Absolute Value Function The absolute value function can be defined as x if x ≥ 0 f ( x= ) x= − x if x < 0 The absolute value function is commonly used to determine the distance between two numbers on the number line. Given two values a and b, then a − b will give the distance, a positive quantity, between these values, regardless of which value is larger. Example 1 Describe all values, x, within a distance of 4 from the number 5. We want the distance between x and 5 to be less than or equal to 4. The distance can be represented using the absolute value, giving the expression x−5 ≤ 4 Example 2 A 2010 poll reported 78% of Americans believe that people who are gay should be able to serve in the US military, with a reported margin of error of 3% 9. The margin of error tells us how far off the actual value could be from the survey value 10. Express the set of possible values using absolute values. Since we want the size of the difference between the actual percentage, p, and the reported percentage to be less than 3%, p − 78 ≤ 3
9
http://www.pollingreport.com/civil.htm, retrieved August 4, 2010 Technically, margin of error usually means that the surveyors are 95% confident that actual value falls within this range. 10
Section 2.5 Absolute Value Functions 147 Try it Now 1. Students who score within 20 points of 80 will pass the test. Write this as a distance from 80 using the absolute value notation. Important Features The most significant feature of the absolute value graph is the corner point where the graph changes direction. When finding the equation for a transformed absolute value function, this point is very helpful for determining the horizontal and vertical shifts. Example 3 Write an equation for the function graphed below.
The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 and down 2 from the basic toolkit function. We might also notice that the graph appears stretched, since the linear portions have slopes of 2 and -2. From this information we can write the equation: f ( x) = 2 x − 3 − 2 , treating the stretch as a vertical stretch
f ( x) = 2( x − 3) − 2 , treating the stretch as a horizontal compression Note that these equations are algebraically equivalent – the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch/compression. If you had not been able to determine the stretch based on the slopes of the lines, you can solve for the stretch factor by putting in a known pair of values for x and f(x) f ( x) = a x − 3 − 2 Now substituting in the point (1, 2)
2 = a1 − 3 − 2 4 = 2a a=2
148 Chapter 2 Try it Now 2. Given the description of the transformed absolute value function write the equation. The absolute value function is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units, The graph of an absolute value function will have a vertical intercept when the input is zero. The graph may or may not have horizontal intercepts, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to have zero, one, or two horizontal intercepts. Zero horizontal intercepts
One
Two
To find the horizontal intercepts, we will need to solve an equation involving an absolute value. Notice that the absolute value function is not one-to-one, so typically inverses of absolute value functions are not discussed. Solving Absolute Value Equations To solve an equation like 8 = 2 x − 6 , we can notice that the absolute value will be equal to eight if the quantity inside the absolute value were 8 or -8. This leads to two different equations we can solve independently: or 2x − 6 = 8 2 x − 6 = −8 2 x = 14 2 x = −2 x=7 x = −1 Solutions to Absolute Value Equations An equation of the form A = B , with B ≥ 0 , will have solutions when A = B or A = − B
Section 2.5 Absolute Value Functions 149 Example 4 Find the horizontal intercepts of the graph of f ( x) = 4 x + 1 − 7 The horizontal intercepts will occur when f ( x) = 0 . Solving, 0 = 4x + 1 − 7 Isolate the absolute value on one side of the equation
7 = 4x + 1 7 = 4x + 1 6 = 4x 6 3 x= = 4 2
Now we can break this into two separate equations: or
− 7 = 4x + 1 − 8 = 4x −8 x= = −2 4
The graph has two horizontal intercepts, at x =
3 and x = -2 2
Example 5 Solve 1 = 4 x − 2 + 2 Isolating the absolute value on one side the equation, 1= 4x−2 +2
−1 = 4 x − 2 −
1 = x−2 4
At this point, we notice that this equation has no solutions – the absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. Try it Now 3. Find the horizontal & vertical intercepts for the function f ( x) = − x + 2 + 3 Solving Absolute Value Inequalities When absolute value inequalities are written to describe a set of values, like the inequality x − 5 ≤ 4 we wrote earlier, it is sometimes desirable to express this set of values without the absolute value, either using inequalities, or using interval notation.
150 Chapter 2 We will explore two approaches to solving absolute value inequalities: 1) Using the graph 2) Using test values Example 6 Solve x − 5 ≤ 4 With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where x − 5 = 4 . We do this because the absolute value is a nice friendly function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve x − 5 = 4 , x−5 = 4 x=9
or
x − 5 = −4 x =1
To use a graph, we can sketch the function f ( x) = x − 5 . To help us see where the outputs are 4, the line g ( x) = 4 could also be sketched.
On the graph, we can see that indeed the output values of the absolute value are equal to 4 at x = 1 and x = 9. Based on the shape of the graph, we can determine the absolute value is less than or equal to 4 between these two points, when 1 ≤ x ≤ 9 . In interval notation, this would be the interval [1,9]. As an alternative to graphing, after determining that the absolute value is equal to 4 at x = 1 and x = 9, we know the graph can only change from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: x 0 As with all inequalities, we start by solving the equality ( x + 3)( x + 1) 2 ( x − 4) = 0 , which has solutions at x = -3, -1, and 4. We know the function can only change from positive to negative at these values, so these divide the inputs into 4 intervals.
180 Chapter 3 We could choose a test value in each interval and evaluate the function f ( x) = ( x + 3)( x + 1) 2 ( x − 4) at each test value to determine if the function is positive or negative in that interval Interval x < -3 -3 < x < -1 -1 < x < 4 x>4
Test x in interval -4 -2 0 5
On a number line this would look like: 0 negative 0 positive
f( test value) 72 -6 -12 288
negative
>0 or 0 4, or in interval notation, (−∞,−3) ∪ (4, ∞) We could have also determined on which intervals the function was positive by sketching a graph of the function. We illustrate that technique in the next example Example 6 Find the domain of the function v(t ) = 6 − 5t − t 2 . A square root is only defined when the quantity we are taking the square root of, the quantity inside the square root, is zero or greater. Thus, the domain of this function will be when 6 − 5t − t 2 ≥ 0 . Again we start by solving the equality 6 − 5t − t 2 = 0 . While we could use the quadratic formula, this equation factors nicely to (6 + t )(1 − t ) = 0 , giving horizontal intercepts t = 1 and t = -6. Sketching a graph of this quadratic will allow us to determine when it is positive. From the graph we can see this function is positive for inputs between the intercepts. So 6 − 5t − t 2 ≥ 0 for − 6 ≤ t ≤ 1 , and this will be the domain of the v(t) function.
3.3 Graphs of Polynomial Functions 181 Try it Now 2. Given the function g ( x) = x 3 − x 2 − 6 x use the methods that we have learned so far to find the vertical & horizontal intercepts, determine where the function is negative and positive, describe the long run behavior and sketch the graph without technology. Writing Equations using Intercepts Since a polynomial function written in factored form will have a horizontal intercept where each factor is equal to zero, we can form a function that will pass through a set of horizontal intercepts by introducing a corresponding set of factors. Factored Form of Polynomials If a polynomial has horizontal intercepts at x = x1 , x 2 , , x n , then the polynomial can be written in the factored form f ( x) = a ( x − x1 ) p1 ( x − x 2 ) p2 ( x − x n ) pn where the powers pi on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the horizontal intercept. Example 7 Write a formula for the polynomial function graphed here.
This graph has three horizontal intercepts: x = -3, 2, and 5. At x = -3 and 5 the graph passes through the axis, suggesting the corresponding factors of the polynomial will be linear. At x = 2 the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be 2nd degree (quadratic). Together, this gives us: f ( x) = a ( x + 3)( x − 2) 2 ( x − 5) To determine the stretch factor, we can utilize another point on the graph. Here, the vertical intercept appears to be (0,-2), so we can plug in those values to solve for a:
182 Chapter 3 − 2 = a (0 + 3)(0 − 2) 2 (0 − 5) − 2 = −60a a=
1 30
The graphed polynomial appears to represent the function 1 f ( x) = ( x + 3)( x − 2) 2 ( x − 5) . 30 Try it Now 3. Given the graph, write a formula for the function shown.
Estimating Extrema With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph. Example 8 An open-top box is to be constructed by cutting out squares from each corner of a 14cm by 20cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. w We will start this problem by drawing a picture, labeling the w width of the cut-out squares with a variable, w.
3.3 Graphs of Polynomial Functions 183 Notice that after a square is cut out from each end, it leaves a (14-2w) cm by (20-2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume: V ( w) = (14 − 2 w)(20 − 2w) w = 280w − 68w 2 + 4w 3 Using technology to sketch a graph allows us to estimate the maximum value for the volume, restricted to reasonable values for w: values from 0 to 7.
From this graph, we can estimate the maximum value is around 340, and occurs when the squares are about 2.75cm square. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph.
From this zoomed-in view, we can refine our estimate for the max volume to about 339, when the squares are 2.7cm square. Try it Now 4. Use technology to find the maximum and minimum values on the interval [-1, 4] of the function f ( x) = −0.2( x − 2) 3 ( x + 1) 2 ( x − 4) .
184 Chapter 3 Important Topics of this Section Short Run Behavior Intercepts (Horizontal & Vertical) Methods to find Horizontal intercepts Factoring Methods Factored Forms Technology Graphical Behavior at intercepts Single, Double and Triple zeros (or power 1, 2, and 3 behaviors) Solving polynomial inequalities using test values & graphing techniques Writing equations using intercepts Estimating extrema Try it Now Answers 1. Vertical intercept (0, 0), Horizontal intercepts (0, 0), (-2, 0), (2, 0) 2. Vertical intercept (0, 0), Horizontal intercepts (-2, 0), (0, 0), (3, 0) The function is negative on ( − ∞ , -2) and (0, 3) The function is positive on (-2, 0) and (3, ∞ ) The leading term is x 3 so as x → −∞ , g (x) → −∞ and as x → ∞ , g (x) → ∞
1 3. f ( x) = − ( x − 2)3 ( x + 1) 2 ( x − 4) 8 4. The minimum occurs at approximately the point (0, -6.5), and the maximum occurs at approximately the point (3.5, 7).
3.3 Graphs of Polynomial Functions 185
Section 3.3 Exercises Find the C and t intercepts of each function. 1. C ( t ) = 2 ( t − 4 )( t + 1) (t − 6) 2. C ( t ) =3 ( t + 2 )( t − 3) (t + 5) 3. C ( t ) =4t ( t − 2 ) (t + 1)
4. C ( t ) = 2t ( t − 3)( t + 1)
5. C ( t ) = 2t 4 − 8t 3 + 6t 2
6. C ( t ) =4t 4 + 12t 3 − 40t 2
2
2
Use your calculator or other graphing technology to solve graphically for the zeros of the function. 7. f ( x ) = x3 − 7 x 2 + 4 x + 30 8. g ( x ) = x3 − 6 x 2 + x + 28 Find the long run behavior of each function as t → ∞ and t → −∞ 2 3 3 3 9. h ( t ) =3 ( t − 5 ) ( t − 3) (t − 2) 10. k ( t ) =2 ( t − 3) ( t + 1) (t + 2) 11. p ( t ) = −2t ( t − 1)( 3 − t )
2
12. q ( t ) = −4t ( 2 − t )( t + 1)
Sketch a graph of each equation. 2 13. f ( x ) =+ ( x 3) ( x − 2)
14. g ( x ) = ( x + 4 )( x − 1)
15. h ( x ) = ( x − 1) ( x + 3)
16. k ( x ) =− ( x 3) ( x − 2 )
3
2
3
3
2
2
−2 x ( x − 1) ( x + 3) 17. m ( x ) =
−3 x ( x + 2 ) ( x − 4) 18. n ( x ) =
Solve each inequality. 2 19. ( x − 3)( x − 2 ) > 0
20. ( x − 5 )( x + 1) > 0
21. ( x − 1)( x + 2 )( x − 3) < 0
22. ( x − 4 )( x + 3)( x + 6 ) < 0
2
Find the domain of each function. 23. f ( x ) = −42 + 19 x − 2 x 2
24. g ( x ) =
28 − 17 x − 3 x 2
25. h ( x ) =
4 − 5x + x2
26. k ( x ) =
2 + 7 x + 3x 2
27. n ( x ) =
( x − 3)( x + 2 )
29. p ( t ) =
1 t + 2t − 8 2
2
28. m ( x ) = ( x − 1) ( x + 3) 2
30. q ( t ) =
4 x − 4x − 5 2
186 Chapter 3 Write an equation for a polynomial the given features. 31. Degree 3. Zeros at x = -2, x = 1, and x = 3. Vertical intercept at (0, -4) 32. Degree 3. Zeros at x = -5, x = -2, and x = 1. Vertical intercept at (0, 6) 33. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = -3. Vertical intercept at (0, 9) 34. Degree 4. Root of multiplicity 2 at x = 4, and a roots of multiplicity 1 at x = 1 and x = -2. Vertical intercept at (0, -3) 35. Degree 5. Double zero at x = 1, and triple zero at x = 3. Passes through the point (2, 15) 36. Degree 5. Single zero at x = -2 and x = 3, and triple zero at x = 1. Passes through the point (2, 4) Write a formula for each polynomial function graphed.
37.
38.
39.
40.
41.
42.
43.
44.
3.3 Graphs of Polynomial Functions 187 Write a formula for each polynomial function graphed.
45.
46.
47.
48.
49.
50.
51. A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y= 5 − x 2 . What are the dimensions of such a rectangle that has the greatest possible area? 52. A rectangle is inscribed with its base on the x axis and its upper corners on the curve = y 16 − x 4 . What are the dimensions of such a rectangle that has the greatest possible area?
188 Chapter 3
Section 3.4 Rational Functions In the last few sections, we have built polynomials based on the positive whole number power functions. In this section we explore functions based on power functions with negative integer powers, called rational functions. Example 1 You plan to drive 100 miles. Find a formula for the time the trip will take as a function of the speed you drive. You may recall that multiplying speed by time will give you distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which we drive, then vt = distance . Since our distance is fixed at 100 miles, vt = 100 . Solving this relationship for the time gives us the function we desired: 100 t (v ) = = 100v −1 v While this type of relationship can be written using the negative exponent, it is more common to see it written as a fraction. This particular example is one of an inversely proportional relationship – where one 5 quantity is a constant divided by the other quantity, like y = . x 1 Notice that this is a transformation of the reciprocal toolkit function, f ( x) = x Several natural phenomena, such as gravitational force and volume of sound, behave in a manner inversely proportional to the square of another quantity. For example, the k volume, V, of a sound heard at a distance d from the source would be related by V = 2 d for some constant value k. 1 These functions are transformations of the reciprocal squared toolkit function f ( x) = 2 . x We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. These graphs have several important features.
3.4 Rational Functions 189
f ( x) =
1 x
f ( x) =
1 x2
1 . As you well know, dividing x by zero is not allowed and therefore zero is not in the domain, and so the function is undefined at an input of zero.
Let’s begin by looking at the reciprocal function, f ( x) =
Short run behavior: As the input values approach zero from the left side (taking on very small, negative values), the function values become very large in the negative direction (in other words, they approach negative infinity). We write: as x → 0 − , f (x) → −∞ . As we approach zero from the right side (small, positive input values), the function values become very large in the positive direction (approaching infinity). We write: as x → 0 + , f (x) → ∞ . This behavior creates a vertical asymptote. An asymptote is a line that the graph approaches. In this case the graph is approaching the vertical line x = 0 as the input becomes close to zero. Long run behavior: As the values of x approach infinity, the function values approach 0. As the values of x approach negative infinity, the function values approach 0. Symbolically: as x → ±∞ , f ( x) → 0 Based on this long run behavior and the graph we can see that the function approaches 0 but never actually reaches 0, it just “levels off” as the inputs become large. This behavior creates a horizontal asymptote. In this case the graph is approaching the horizontal line f ( x) = 0 as the input becomes very large in the negative and positive directions. Vertical and Horizontal Asymptotes A vertical asymptote of a graph is a vertical line x = a where the graph tends towards positive or negative infinity as the inputs approach a. As x → a , f (x) → ±∞ . A horizontal asymptote of a graph is a horizontal line y = b where the graph approaches the line as the inputs get large. As x → ±∞ , f ( x) → b .
190 Chapter 3 Try it Now: 1. Use symbolic notation to describe the long run behavior and short run behavior for the reciprocal squared function.
Example 2 Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any. Transforming the graph left 2 and up 3 would result in the function 1 f ( x) = + 3 , or equivalently, by giving the terms a common denominator, x+2 3x + 7 f ( x) = x+2 Shifting the toolkit function would give us this graph. Notice that this equation is undefined at x = -2, and the graph also is showing a vertical asymptote at x = -2. As x → −2− , f ( x) → −∞ , and as x → −2+ , f ( x) → ∞ As the inputs grow large, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y = 3 . As x → ±∞ , f ( x) → 3 . Notice that horizontal and vertical asymptotes get shifted left 2 and up 3 along with the function. Try it Now 2. Sketch the graph and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units. In the previous example, we shifted a toolkit function in a way that resulted in a function 3x + 7 of the form f ( x) = . This is an example of a more general rational function. x+2
3.4 Rational Functions 191 Rational Function A rational function is a function that can be written as the ratio of two polynomials, P(x) and Q(x). 2 p P( x) a0 + a1 x + a2 x + + a p x = f ( x) = Q( x) b0 + b1 x + b2 x 2 + + bq x q Example 3 A large mixing tank currently contains 100 gallons of water, into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. Notice that the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each: water = 100 + 10t sugar = 5 + 1t The concentration, C, will be the ratio of pounds of sugar to gallons of water 5+t C (t ) = 100 + 10t Finding Asymptotes and Intercepts Given a rational function, as part of investigating the short run behavior we are interested in finding any vertical and horizontal asymptotes, as well as finding any vertical or horizontal intercepts, as we have done in the past. To find vertical asymptotes, we notice that the vertical asymptotes in our examples occur when the denominator of the function is undefined. With one exception, a vertical asymptote will occur whenever the denominator is undefined. Example 4 5 + 2x 2 Find the vertical asymptotes of the function k ( x) = 2 − x − x2 To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: 2 − x − x2 = 0 (2 + x)(1 − x) = 0 x = −2, 1
192 Chapter 3 This indicates two vertical asymptotes, which a look at a graph confirms.
The exception to this rule can occur when both the numerator and denominator of a rational function are zero at the same input. Example 5 Find the vertical asymptotes of the function k ( x) =
x−2 . x2 − 4
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: 0 x2 − 4 =
x2 = 4 x = −2, 2 However, the numerator of this function is also equal to zero when x = 2. Because of this, the 0 function will still be undefined at 2, since is 0 undefined, but the graph will not have a vertical asymptote at x = 2. The graph of this function will have the vertical asymptote at x = -2, but at x = 2 the graph will have a hole: a single point where the graph is not defined, indicated by an open circle. Vertical Asymptotes and Holes of Rational Functions The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. A hole might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. In this case, factor the numerator and denominator and simplify; if the simplified expression still has a zero in the denominator at the original input the original function has a vertical asymptote at the input, otherwise it has a hole.
3.4 Rational Functions 193 To find horizontal asymptotes, we are interested in the behavior of the function as the input grows large, so we consider long run behavior of the numerator and denominator separately. Recall that a polynomial’s long run behavior will mirror that of the leading term. Likewise, a rational function’s long run behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. There are three distinct outcomes when this analysis is done: Case 1: The degree of the denominator > degree of the numerator 3x + 2 Example: f ( x) = 2 x + 4x − 5 3x 3 . This tells us that as the inputs grow In this case, the long run behavior is f ( x) ≈ 2 = x x 3 large, this function will behave similarly to the function g ( x) = . As the inputs grow x large, the outputs will approach zero, resulting in a horizontal asymptote at y = 0 . As x → ±∞ , f ( x) → 0 Case 2: The degree of the denominator < degree of the numerator 3x 2 + 2 Example: f ( x) = x−5 3x 2 In this case, the long run behavior is f ( x) ≈ = 3 x . This tells us that as the inputs x grow large, this function will behave similarly to the function g ( x) = 3 x . As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. As x → ±∞ , f (x) → ±∞ , respectively. Ultimately, if the numerator is larger than the denominator, the long run behavior of the graph will mimic the behavior of the reduced long run behavior fraction. As another 3x5 − x 2 example if we had the function f ( x) = with long run behavior x+3 3x5 f ( x) ≈ = 3 x 4 , the long run behavior of the graph would look similar to that of an x even polynomial, and as x → ±∞ , f (x) → ∞ . Case 3: The degree of the denominator = degree of the numerator 3x 2 + 2 Example: f ( x) = 2 x + 4x − 5
194 Chapter 3 3x 2 = 3 . This tells us that as the inputs grow x2 large, this function will behave like the function g ( x) = 3 , which is a horizontal line. As x → ±∞ , f ( x) → 3 , resulting in a horizontal asymptote at y = 3 . In this case, the long run behavior is f ( x) ≈
Horizontal Asymptote of Rational Functions The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. Degree of denominator > degree of numerator: Horizontal asymptote at y = 0 Degree of denominator < degree of numerator: No horizontal asymptote Degree of denominator = degree of numerator: Horizontal asymptote at ratio of leading coefficients. Example 6 In the sugar concentration problem from earlier, we created the equation 5+t . C (t ) = 100 + 10t Find the horizontal asymptote and interpret it in context of the scenario. Both the numerator and denominator are linear (degree 1), so since the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these 1 values: As t → ∞ , C (t ) → . This function will have a horizontal asymptote at 10 1 y= . 10 This tells us that as the input gets large, the output values will approach 1/10. In context, this means that as more time goes by, the concentration of sugar in the tank will approach one tenth of a pound of sugar per gallon of water or 1/10 pounds per gallon. Example 7 Find the horizontal and vertical asymptotes of the function ( x − 2)( x + 3) f ( x) = ( x − 1)( x + 2)( x − 5) First, note this function has no inputs that make both the numerator and denominator zero, so there are no potential holes. The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x = 1, -2, and 5, indicating vertical asymptotes at these values.
3.4 Rational Functions 195 The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as x → ±∞ , f ( x) → 0 . This function will have a horizontal asymptote at y = 0. Try it Now 3. Find the vertical and horizontal asymptotes of the function (2 x − 1)(2 x + 1) f ( x) = ( x − 2)( x + 3) Intercepts As with all functions, a rational function will have a vertical intercept when the input is zero, if the function is defined at zero. It is possible for a rational function to not have a vertical intercept if the function is undefined at zero. Likewise, a rational function will have horizontal intercepts at the inputs that cause the output to be zero (unless that input corresponds to a hole). It is possible there are no horizontal intercepts. Since a fraction is only equal to zero when the numerator is zero, horizontal intercepts will occur when the numerator of the rational function is equal to zero. Example 8 Find the intercepts of f ( x) =
( x − 2)( x + 3) ( x − 1)( x + 2)( x − 5)
We can find the vertical intercept by evaluating the function at zero −6 (0 − 2)(0 + 3) 3 f (0) = = =− (0 − 1)(0 + 2)(0 − 5) 10 5 The horizontal intercepts will occur when the function is equal to zero: ( x − 2)( x + 3) This is zero when the numerator is zero 0= ( x − 1)( x + 2)( x − 5) 0 = ( x − 2)( x + 3) x = 2, − 3 Try it Now 4. Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function and find the horizontal and vertical intercepts and the horizontal and vertical asymptotes.
196 Chapter 3 From the previous example, you probably noticed that the numerator of a rational function reveals the horizontal intercepts of the graph, while the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Happily, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials. When factors of the denominator have integer powers greater than one, the behavior at the corresponding vertical asymptote will mirror one of the two toolkit reciprocal functions. We get this behavior when the degree of the factor in the denominator is odd. The distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity.
We get this behavior when the degree of the factor in the denominator is even. The distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote, or heads toward negative infinity on both sides.
For example, the graph of ( x + 1) 2 ( x − 3) is shown here. f ( x) = ( x + 3) 2 ( x − 2) At the horizontal intercept x = -1 corresponding to the ( x + 1) 2 factor of the numerator, the graph bounces at the intercept, consistent with the quadratic nature of the factor. At the horizontal intercept x = 3 corresponding to the ( x − 3) factor of the numerator, the graph passes through the axis as we’d expect from a linear factor. At the vertical asymptote x = -3 corresponding to the ( x + 3) 2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with 1 the behavior of the 2 toolkit. x
3.4 Rational Functions 197 At the vertical asymptote x = 2 corresponding to the ( x − 2) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards 1 negative infinity on the right side, consistent with the behavior of the toolkit. x Example 9 Sketch a graph of f ( x) =
( x + 2)( x − 3) . ( x + 1) 2 ( x − 2)
We can start our sketch by finding intercepts and asymptotes. Evaluating the function at zero gives the vertical intercept: (0 + 2)(0 − 3) = f (0) = 3 (0 + 1) 2 (0 − 2) Looking at when the numerator of the function is zero, we can determine the graph will have horizontal intercepts at x = -2 and x = 3. At each, the behavior will be linear, with the graph passing through the intercept. Looking at when the denominator of the function is zero, we can determine the graph will have vertical asymptotes at x = -1 and x = 2. Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. To sketch the graph, we might start by plotting the three intercepts. Since the graph has no horizontal intercepts between the vertical asymptotes, and the vertical intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph. Since the factor associated with the vertical asymptote at x = -1 was squared, we know the graph will have the same behavior on both sides of the asymptote. Since the graph heads towards positive infinity as the inputs approach the asymptote on the right, the graph will head towards positive infinity on the left as well. For the vertical asymptote at x = 2, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. After passing through the horizontal intercepts, the graph will then level off towards an output of zero, as indicated by the horizontal asymptote.
198 Chapter 3 Try it Now ( x + 2) 2 ( x − 2) , use the characteristics of polynomials 2( x − 1) 2 ( x − 3) and rational functions to describe its behavior and sketch the function . 5. Given the function f ( x) =
Since a rational function written in factored form will have a horizontal intercept where each factor of the numerator is equal to zero, we can form a numerator that will pass through a set of horizontal intercepts by introducing a corresponding set of factors. Likewise, since the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors. Writing Rational Functions from Intercepts and Asymptotes If a rational function has horizontal intercepts at x = x1 , x 2 , , x n , and vertical asymptotes at x = v1 , v 2 , , v m then the function can be written in the form ( x − x1 ) p1 ( x − x 2 ) p2 ( x − x n ) pn f ( x) = a ( x − v1 ) q1 ( x − v 2 ) q2 ( x − v m ) qn where the powers pi or qi on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the horizontal intercept, or by the horizontal asymptote if it is nonzero. Example 10 Write an equation for the rational function graphed here. The graph appears to have horizontal intercepts at x = -2 and x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = -1 seems to exhibit the basic 1 behavior similar to , with the graph heading toward positive infinity on one side and x heading toward negative infinity on the other. The asymptote at x = 2 is exhibiting a 1 behavior similar to 2 , with the graph heading toward negative infinity on both sides x of the asymptote.
3.4 Rational Functions 199 Utilizing this information indicates an function of the form ( x + 2)( x − 3) f ( x) = a ( x + 1)( x − 2) 2 To find the stretch factor, we can use another clear point on the graph, such as the vertical intercept (0,-2): (0 + 2)(0 − 3) −2= a (0 + 1)(0 − 2) 2 −6 −2= a 4 −8 4 a= = −6 3 This gives us a final function of f ( x) =
4( x + 2)( x − 3) 3( x + 1)( x − 2) 2
Important Topics of this Section Inversely proportional; Reciprocal toolkit function Inversely proportional to the square; Reciprocal squared toolkit function Horizontal Asymptotes Vertical Asymptotes Rational Functions Finding intercepts, asymptotes, and holes. Given equation sketch the graph Identifying a function from its graph Try it Now Answers 1. Long run behavior, as x → ±∞ , f ( x) → 0 Short run behavior, as x → 0 , f (x) → ∞ (there are no horizontal or vertical intercepts) 2.
The function and the asymptotes are shifted 3 units right and 4 units down. As x → 3 , f (x) → ∞ and as x → ±∞ , f ( x) → −4
200 Chapter 3
3. Vertical asymptotes at x = 2 and x = -3; horizontal asymptote at y = 4 4. For the transformed reciprocal squared function, we find the rational form. 1 1 − 4( x − 3) 2 1 − 4( x 2 − 6 x + 9) − 4 x 2 + 24 x − 35 f ( x) = −4= = = ( x − 3)( x − 3) x 2 − 6x + 9 ( x − 3) 2 ( x − 3) 2 Since the numerator is the same degree as the denominator we know that as x → ±∞ , f ( x) → −4 . y = −4 is the horizontal asymptote. Next, we set the denominator equal to zero to find the vertical asymptote at x = 3, because as x → 3 , f (x) → ∞ . We set the numerator equal to 0 and find the horizontal intercepts are at − 35 (2.5,0) and (3.5,0), then we evaluate at 0 and the vertical intercept is at 0, 9 5. Horizontal asymptote at y = 1/2. Vertical asymptotes are at x = 1, and x = 3. Vertical intercept at (0, 4/3), Horizontal intercepts (2, 0) and (-2, 0) (-2, 0) is a double zero and the graph bounces off the axis at this point. (2, 0) is a single zero and crosses the axis at this point.
3.4 Rational Functions 201
Section 3.4 Exercises Match each equation form with one of the graphs. x− A 1. f ( x ) = x−B
2. g ( x )
A
B
( x − A) = x−B
2
3. h ( x ) =
( x − B)
( x − A) k ( x) = 2 ( x − B) 2
x− A
4.
2
C
D
For each function, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal asymptote. Use that information to sketch a graph. 5. p ( x ) = 7. s ( x ) =
2x − 3 x+4
4
( x − 2)
2
6. q ( x ) =
x −5 3x − 1
8. r ( x ) =
5
( x + 1)
2
3 x 2 − 14 x − 5 3 x 2 + 8 x − 16
10. g ( x ) =
2 x 2 + 7 x − 15 3 x 2 − 14 + 15
11. a ( x ) =
x2 + 2x − 3 x2 −1
12. b ( x ) =
x2 − x − 6 x2 − 4
13. h ( x ) =
2 x 2 + x − 1 x−4
14. k ( x ) =
2 x 2 − 3 x − 20 x −5
15. n ( x ) =
3x 2 + 4 x − 4 x3 − 4 x 2
16. m ( x ) =
5− x 2x + 7x + 3
9. f ( x ) =
17.
( x − 1)( x + 3)( x − 5) w( x) = 2 ( x + 2 ) ( x − 4)
2
( x + 2 ) ( x − 5) z ( x) = ( x − 3)( x + 1)( x + 4 ) 2
18.
202 Chapter 3 Write an equation for a rational function with the given characteristics. 19. Vertical asymptotes at x = 5 and x = −5 x intercepts at (2, 0) and (−1, 0)
y intercept at ( 0, 4 )
20. Vertical asymptotes at x = −4 and x = −1 x intercepts at (1, 0 ) and ( 5, 0 ) y intercept at (0, 7) 21. Vertical asymptotes at x = −4 and x = −5 x intercepts at ( 4, 0 ) and ( −6, 0 ) Horizontal asymptote at y = 7 22. Vertical asymptotes at x = −3 and x = 6 x intercepts at ( −2, 0 ) and (1, 0 )
Horizontal asymptote at y = −2
23. Vertical asymptote at x = −1 Double zero at x = 2
y intercept at (0, 2)
24. Vertical asymptote at x = 3 Double zero at x = 1
y intercept at (0, 4)
Write an equation for the function graphed.
25.
27.
26.
28.
\
3.4 Rational Functions 203 Write an equation for the function graphed.
29.
30.
31.
32.
33.
34.
35.
36.
204 Chapter 3 Write an equation for the function graphed.
37.
38.
39. A scientist has a beaker containing 20 mL of a solution containing 20% acid. To dilute this, she adds pure water. a. Write an equation for the concentration in the beaker after adding n mL of water. b. Find the concentration if 10 mL of water has been added. c. How many mL of water must be added to obtain a 4% solution? d. What is the behavior as n → ∞ , and what is the physical significance of this? 40. A scientist has a beaker containing 30 mL of a solution containing 3 grams of potassium hydroxide. To this, she mixes a solution containing 8 milligrams per mL of potassium hydroxide. a. Write an equation for the concentration in the tank after adding n mL of the second solution. b. Find the concentration if 10 mL of the second solution has been added. c. How many mL of water must be added to obtain a 50 mg/mL solution? d. What is the behavior as n → ∞ , and what is the physical significance of this? 41. Oscar is hunting magnetic fields with his gauss meter, a device for measuring the strength and polarity of magnetic fields. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. [UW] ax + b relating the meter reading m( x) a. Give a rational model of form m ( x ) = cx + d to how many feet x Oscar has gone into the room. b. How far must he go for the meter to reach 10? 100? c. Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is?
3.4 Rational Functions 205 42. The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%. You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score, p (n) , is a function of the number of hours, n, that you study in the form p (n) = you need to study? [UW]
an + b . If you want a score of 80%, how long do cn + d
43. A street light is 10 feet north of a straight bike path that runs eastwest. Olav is bicycling down the path at a rate of 15 miles per hour. At noon, Olav is 33 feet west of the point on the bike path closest to the street light. (See the picture). The relationship between the intensity C of light (in candlepower) and the k distance d (in feet) from the light source is given by C = 2 , where k is a constant d depending on the light source. [UW] a. From 20 feet away, the street light has an intensity of 1 candle. What is k? b. Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds. c. When will the light on Olav have maximum intensity? d. When will the intensity of the light be 2 candles?
206 Chapter 3
Section 3.5 Inverses and Radical Functions In this section, we will explore the inverses of polynomial and rational functions, and in particular the radical functions that arise in the process. Example 1 A water runoff collector is built in the shape of a parabolic trough as shown below. Find the surface area of the water in the trough as a function of the depth of the water. 3ft
12 in
18 in
Since it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x measured horizontally and y measured vertically, with the origin at the vertex of the parabola. y
x
From this we find an equation for the parabolic shape. Since we placed the origin at the vertex of the parabola, we know the equation will have form y ( x) = ax 2 . Our equation will need to pass through the point (6,18), from which we can solve for the stretch factor a: 18 = a 6 2 18 1 = a= 36 2 1 Our parabolic cross section has equation y ( x) = x 2 2 Since we are interested in the surface area of the water, we are interested in determining the width at the top of the water as a function of the water depth. For any depth y the width will be given by 2x, so we need to solve the equation above for x. However notice that the original function is not one-to-one, and indeed given any output there are two inputs that produce the same output, one positive and one negative.
3.5 Inverses and Radical Functions 207 To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable: 1 y = x2 2 2y = x2
x = ± 2y This is not a function as written. Since we are limiting ourselves to positive x values, we eliminate the negative solution, giving us the inverse function we’re looking for x( y ) = 2 y Since x measures from the center out, the entire width of the water at the top will be 2x. Since the trough is 3 feet (36 inches) long, the surface area will then be 36(2x), or in terms of y: Area = 72 x = 72 2 y The previous example illustrated two important things: 1) When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one. 2) The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions. Functions involving roots are often called radical functions. Example 2 Find the inverse of f ( x) = ( x − 2) 2 − 3 = x 2 − 4 x + 1 From the transformation form of the function, we can see this is a transformed quadratic with vertex at (2,-3) that opens upwards. Since the graph will be decreasing on one side of the vertex, and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to x ≥ 2 . To find the inverse, we will use the vertex form of the quadratic. We start by replacing the f(x) with a simple variable y, then solve for x. Add 3 to both sides y = ( x − 2) 2 − 3 2 Take the square root y + 3 = ( x − 2) ±
y+3=x−2
2± y+3 = x
Add 2 to both sides
208 Chapter 3 Of course, as written this is not a function. Since we restricted our original function to a domain of x ≥ 2 , the outputs of the inverse should be the same, telling us to utilize the + case: x = f −1 ( y ) = 2 + y + 3 If the quadratic had not been given in vertex form, rewriting it into vertex form is probably the best approach. Alternatively, we could have taken the standard equation and rewritten it equal to zero: 0 = x 2 − 4x + 1 − y We would then be able to use the quadratic formula with a = 1 , b = −4 , and c= (1 − y ) , resulting in the same solutions we found above: −(−4) ± (−4) 2 − 4(1)(1 − y ) 12 + 4 y x= =2 ± =2 ± 3 + y 2 2 Try it Now 1. Find the inverse of the function f ( x= ) x 2 + 1 , on the domain x ≥ 0 . While it is not possible to find an inverse of most polynomial functions, some other basic polynomials are invertible. Example 3 Find the inverse of the function f ( x) = 5 x 3 + 1 . This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x y = 5x 3 + 1 y − 1 = 5x 3 y −1 = x3 5 x = f −1 ( y ) = 3
y −1 5
Notice that this inverse is also a transformation of a power function with a fractional power, x1/3. Try it Now 2. Which toolkit functions have inverse functions without restricting their domain?
3.5 Inverses and Radical Functions 209 Besides being important as an inverse function, radical functions are common in important physical models. Example 4 The velocity, v in feet per second, of a car that slammed on its brakes can be determined based on the length of skid marks that the tires left on the ground. This relationship is given by v(d ) = 2 gfd In this formula, g represents acceleration due to gravity (32 ft/sec2), d is the length of the skid marks in feet, and f is a constant representing the friction of the surface. A car lost control on wet asphalt, with a friction coefficient of 0.5, leaving 200 foot skid marks. How fast was the car travelling when it lost control? Using the given values of f = 0.5 and d = 200, we can evaluate the given formula: = v(200) = 2(32)(0.5)(200) 80 ft/sec , which is about 54.5 miles per hour. When radical functions are composed with other functions, determining domain can become more complicated. Example 5 Find the domain of the function f ( x) =
( x + 2)( x − 3) . ( x − 1)
Since a square root is only defined when the quantity under the radical is non-negative, ( x + 2)( x − 3) we need to determine where ≥ 0 . A rational function can change signs ( x − 1) (change from positive to negative or vice versa) at horizontal intercepts and at vertical asymptotes. For this equation, the graph could change signs at x = -2, 1, and 3. To determine on which intervals the rational expression is positive, we could evaluate the expression at test values, or sketch a graph. While both approaches work equally well, for this example we will use a graph. This function has two horizontal intercepts, both of which exhibit linear behavior, where the graph will pass through the intercept. There is one vertical asymptote, corresponding to a linear factor, leading to a behavior similar to the basic reciprocal toolkit function. There is a vertical intercept at (0, 6). This graph does not have a horizontal asymptote, since the degree of the numerator is larger than the degree of the denominator.
210 Chapter 3 From the vertical intercept and horizontal intercept at x = -2, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph.
From the graph, we can now tell on which intervals this expression will be non-negative, so the original function f(x) will be defined. f(x) has domain − 2 ≤ x < 1 or x ≥ 3 , or in interval notation, [−2,1) ∪ [3, ∞) . Like with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as our concentration examples. Example 6
20 + 0.4n was used in the previous section to represent the 100 + n concentration of an acid solution after n mL of 40% solution has been added to 100 mL of a 20% solution. We might want to be able to determine instead how much 40% solution has been added based on the current concentration of the mixture.
The function C (n) =
To do this, we would want the inverse of this function: 20 + 0.4n multiply both sides by the denominator C= 100 + n distribute C (100 + n) = 20 + 0.4n group everything with n on one side 100C + Cn = 20 + 0.4n factor out n 100C − 20 = 0.4n − Cn divide to find the inverse 100C − 20 = (0.4 − C )n 100C − 20 n(C ) = 0.4 − C If, for example, we wanted to know how many mL of 40% solution need to be added to obtain a concentration of 35%, we can simply evaluate the inverse rather than solving an equation involving the original function: 100(0.35) − 20 15 n(0.35) = = = 300 mL of 40% solution would need to be added. 0.4 − 0.35 0.05 Try it Now 3. Find the inverse of the function f ( x) =
x+3 . x−2
3.5 Inverses and Radical Functions 211 Important Topics of this Section Imposing a coordinate system Finding an inverse function Restricting the domain Invertible toolkit functions Radical Functions Inverses of rational functions Try it Now Answers 1. x = f −1 ( y ) = y − 1 2. identity, cubic, square root, cube root 2y + 3 3. f −1 ( y ) = y −1
212 Chapter 3
Section 3.5 Exercises For each function, find a domain on which the function is one-to-one and non-decreasing, then find an inverse of the function on this domain. 2 2 1. f ( x= 2. f ( x= ) ( x − 4) ) ( x + 2) 3. f ( x= ) 12 − x 2
4. f ( x )= 9 − x 2
5. f (= x ) 3x3 + 1
6. f ( x )= 4 − 2 x 3
Find the inverse of each function. 7. f ( x ) = 9 + 4x − 4
8. f ( x= )
9. f ( x )= 9 + 2 3 x
10. f ( x )= 3 − 3 x
6x − 8 + 5
11. f ( x ) =
2 x +8
12. f ( x ) =
3 x−4
13. f ( x ) =
x+3 x+7
14. f ( x ) =
x−2 x+7
15. f ( x ) =
3x + 4 5 − 4x
16. f ( x ) =
5x + 1 2 − 5x
Police use the formula v = 20 L to estimate the speed of a car, v, in miles per hour, based on the length, L, in feet, of its skid marks when suddenly braking on a dry, asphalt road. 17. At the scene of an accident, a police officer measures a car's skid marks to be 215 feet long. Approximately how fast was the car traveling? 18. At the scene of an accident, a police officer measures a car's skid marks to be 135 feet long. Approximately how fast was the car traveling? The formula v = 2.7 r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r, in feet. 19. A highway crew measures the radius of curvature at an exit ramp on a highway as 430 feet. What is the maximum safe speed? 20. A highway crew measures the radius of curvature at a tight corner on a highway as 900 feet. What is the maximum safe speed?
3.5 Inverses and Radical Functions 213 21. A drainage canal has a crosssection in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch? [UW]
22. Brooke is located 5 miles out from the nearest point A along a straight shoreline in her sea kayak. Hunger strikes and she wants to make it to Kono’s for lunch; see picture. Brooke can paddle 2 mph and walk 4 mph. [UW] a. If she paddles along a straight line course to the shore, find an expression that computes the total time to reach lunch in terms of the location where Brooke beaches her kayak. b. Determine the total time to reach Kono’s if she paddles directly to the point A. c. Determine the total time to reach Kono’s if she paddles directly to Kono’s. d. Do you think your answer to b or c is the minimum time required for Brooke to reach lunch? e. Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point A and Kono’s. How does this time compare to the times in parts b or c? Do you need to modify your answer to part d? 23. Clovis is standing at the edge of a dropoff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing. With the origin of the coordinate system located where he is standing, and the x-axis extending horizontally, the path of the rocket is described by the formula y= −2 x 2 + 120 x . [UW] a. Give a function h = f ( x) relating the height h of the rocket above the sloping ground to its x-coordinate. b. Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height? c. Clovis measures the height h of the rocket above the sloping ground while it is going up. Give a function x = g ( h ) relating the x-coordinate of the rocket to h. d. Does the function from (c) still work when the rocket is going down? Explain.
214 Chapter 3
24. A trough has a semicircular cross section with a radius of 5 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. [UW] a. Give a function w = f ( t ) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too. b. After how many hours will the surface of the water have width of 6 feet? c. Give a function t = f −1 ( w ) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too.
Chapter 4: Exponential and Logarithmic Functions
Section 4.1 Exponential Functions ............................................................................. 215 Section 4.2 Graphs of Exponential Functions............................................................. 232 Section 4.3 Logarithmic Functions ............................................................................. 242 Section 4.4 Logarithmic Properties............................................................................. 253 Section 4.5 Graphs of Logarithmic Functions ............................................................ 262 Section 4.6 Exponential and Logarithmic Models ...................................................... 270 Section 4.7 Fitting Exponentials to Data .................................................................... 289
Section 4.1 Exponential Functions India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year 1. We might ask if we can find a formula to model the population, P, as a function of time, t, in years after 2008, if the population continues to grow at this rate. In linear growth, we had a constant rate of change – a constant number that the output increased for each increase in input. For example, in the equation f ( x) = 3 x + 4 , the slope tells us the output increases by three each time the input increases by one. This population scenario is different – we have a percent rate of change rather than a constant number of people as our rate of change. To see the significance of this difference consider these two companies: Company A has 100 stores, and expands by opening 50 new stores a year Company B has 100 stores, and expands by increasing the number of stores by 50% of their total each year. Looking at a few years of growth for these companies: Year 0
Stores, company A 100
Starting with 100 each
Stores, company B 100
1
100 + 50 = 150
They both grow by 50 stores in the first year.
100 + 50% of 100 100 + 0.50(100) = 150
2
150 + 50 = 200
Store A grows by 50, Store B grows by 75
150 + 50% of 150 150 + 0.50(150) = 225
3
200 + 50 = 250
Store A grows by 50, Store B grows by 112.5
225 + 50% of 225 225 + 0.50(225) = 337.5
1
World Bank, World Development Indicators, as reported on http://www.google.com/publicdata, retrieved August 20, 2010 This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
216 Chapter 4 Notice that with the percent growth, each year the company is grows by 50% of the current year’s total, so as the company grows larger, the number of stores added in a year grows as well. To try to simplify the calculations, notice that after 1 year the number of stores for company B was: or equivalently by factoring 100 + 0.50(100) 100(1 + 0.50) = 150 We can think of this as “the new number of stores is the original 100% plus another 50%”. After 2 years, the number of stores was: or equivalently by factoring 150 + 0.50(150) now recall the 150 came from 100(1+0.50). Substituting that, 150(1 + 0.50) 100(1 + 0.50)(1 + 0.50) = 100(1 + 0.50) 2 = 225 After 3 years, the number of stores was: or equivalently by factoring 225 + 0.50(225) now recall the 225 came from 100(1 + 0.50) 2 . Substituting that, 225(1 + 0.50) 100(1 + 0.50) 2 (1 + 0.50) = 100(1 + 0.50) 3 = 337.5 From this, we can generalize, noticing that to show a 50% increase, each year we multiply by a factor of (1+0.50), so after n years, our equation would be B(n) = 100(1 + 0.50) n In this equation, the 100 represented the initial quantity, and the 0.50 was the percent growth rate. Generalizing further, we arrive at the general form of exponential functions. Exponential Function An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. The equation can be written in the form f ( x) = a (1 + r ) x or f ( x) = ab x where b = 1+r Where a is the initial or starting value of the function r is the percent growth or decay rate, written as a decimal b is the growth factor or growth multiplier. Since powers of negative numbers behave strangely, we limit b to positive values. To see more clearly the difference between exponential and linear growth, compare the two tables and graphs below, which illustrate the growth of company A and B described above over a longer time frame if the growth patterns were to continue
Section 4.1 Exponential Functions 217
years
2 4 6 8 10
Company A Company B 200 225 300 506 400 1139 500 2563 600 5767
B
A
Example 1 Write an exponential function for India’s population, and use it to predict the population in 2020. At the beginning of the chapter we were given India’s population of 1.14 billion in the year 2008 and a percent growth rate of 1.34%. Using 2008 as our starting time (t = 0), our initial population will be 1.14 billion. Since the percent growth rate was 1.34%, our value for r is 0.0134. Using the basic formula for exponential growth f ( x) = a (1 + r ) x we can write the formula, f (t ) = 1.14(1 + 0.0134) t To estimate the population in 2020, we evaluate the function at t = 12, since 2020 is 12 years after 2008. f (12) = 1.14(1 + 0.0134)12 ≈ 1.337 billion people in 2020 Try it Now 1. Given the three statements below, identify which represent exponential functions. A. The cost of living allowance for state employees increases salaries by 3.1% each year. B. State employees can expect a $300 raise each year they work for the state. C. Tuition costs have increased by 2.8% each year for the last 3 years. Example 2 A certificate of deposit (CD) is a type of savings account offered by banks, typically offering a higher interest rate in return for a fixed length of time you will leave your money invested. If a bank offers a 24 month CD with an annual interest rate of 1.2% compounded monthly, how much will a $1000 investment grow to over those 24 months? First, we must notice that the interest rate is an annual rate, but is compounded monthly, meaning interest is calculated and added to the account monthly. To find the monthly interest rate, we divide the annual rate of 1.2% by 12 since there are 12 months in a
218 Chapter 4 year: 1.2%/12 = 0.1%. Each month we will earn 0.1% interest. From this, we can set up an exponential function, with our initial amount of $1000 and a growth rate of r = 0.001, and our input m measured in months. m
.012 f (m) = 10001 + 12 f (m) = 1000(1 + 0.001) m After 24 months, the account will have grown to f (24) = 1000(1 + 0.001) 24 =$1024.28
Try it Now 2. Looking at these two equations that represent the balance in two different savings accounts, which account is growing faster, and which account will have a higher balance after 3 years? t t A(t ) = 1000(1.05) B(t ) = 900(1.075) In all the preceding examples, we saw exponential growth. Exponential functions can also be used to model quantities that are decreasing at a constant percent rate. An example of this is radioactive decay, a process in which radioactive isotopes of certain atoms transform to an atom of a different type, causing a percentage decrease of the original material over time. Example 3 Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one week? With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is a = 100 mg, and our growth rate will be negative 13%, since we are decreasing: r = -0.13. This gives the equation: Q(d ) = 100(1 − 0.13) d = 100(0.87) d This can also be explained by recognizing that if 13% decays, then 87 % remains. After one week, 7 days, the quantity remaining would be Q(7) = 100(0.87) 7 = 37.73 mg of Bismuth-210 remains. Try it Now 3. A population of 1000 is decreasing 3% each year. Find the population in 30 years.
Section 4.1 Exponential Functions 219 Example 4 T(q) represents the total number of Android smart phone contracts, in thousands, held by a certain Verizon store region measured quarterly since January 1, 2010, Interpret all of the parts of the equation T (2) = 86(1.64) 2 = 231.3056 . Interpreting this from the basic exponential form, we know that 86 is our initial value. This means that on Jan. 1, 2010 this region had 86,000 Android smart phone contracts. Since b = 1 + r = 1.64, we know that every quarter the number of smart phone contracts grows by 64%. T(2) = 231.3056 means that in the 2nd quarter (or at the end of the second quarter) there were approximately 231,305 Android smart phone contracts. Finding Equations of Exponential Functions In the previous examples, we were able to write equations for exponential functions since we knew the initial quantity and the growth rate. If we do not know the growth rate, but instead know only some input and output pairs of values, we can still construct an exponential function. Example 5 In 2002, 80 deer were reintroduced into a wildlife refuge area from which the population had previously been hunted to elimination. By 2008, the population had grown to 180 deer. If this population grows exponentially, find a formula for the function. By defining our input variable to be t, years after 2002, the information listed can be written as two input-output pairs: (0,80) and (6,180). Notice that by choosing our input variable to be measured as years after the first year value provided, we have effectively “given” ourselves the initial value for the function: a = 80. This gives us an equation of the form f (t ) = 80b t . Substituting in our second input-output pair allows us to solve for b: Divide by 80 180 = 80b 6 180 9 Take the 6th root of both sides. = b6 = 80 4 9 = b 6= 1.1447 4 This gives us our equation for the population: f (t ) = 80(1.1447) t Recall that since b = 1+r, we can interpret this to mean that the population growth rate is r = 0.1447, and so the population is growing by about 14.47% each year. In this example, you could also have used (9/4)^(1/6) to evaluate the 6th root if your calculator doesn’t have an nth root button.
220 Chapter 4 In the previous example, we chose to use the f ( x) = ab x form of the exponential function rather than the f ( x) = a (1 + r ) x form. This choice was entirely arbitrary – either form would be fine to use. When finding equations, the value for b or r will usually have to be rounded to be written easily. To preserve accuracy, it is important to not over-round these values. Typically, you want to be sure to preserve at least 3 significant digits in the growth rate. For example, if your value for b was 1.00317643, you would want to round this no further than to 1.00318. In the previous example, we were able to “give” ourselves the initial value by clever definition of our input variable. Next we consider a situation where we can’t do this. Example 6 Find a formula for an exponential function passing through the points (-2,6) and (2,1). Since we don’t have the initial value, we will take a general approach that will work for any function form with unknown parameters: we will substitute in both given inputoutput pairs in the function form f ( x) = ab x and solve for the unknown values, a and b. Substituting in (-2, 6) gives 6 = ab −2 Substituting in (2, 1) gives 1 = ab 2 We now solve these as a system of equations. To do so, we could try a substitution approach, solving one equation for a variable, then substituting that expression into the second equation. Solving 6 = ab −2 for a: 6 a = 6b 2 = −2 b In the second equation, 1 = ab 2 , we substitute the expression above for a: 1 = (6b 2 )b 2 1 = 6b 4 1 = b4 6 1 b = 4 ≈ 0.6389 6 Going back to the equation a = 6b 2 lets us find a: a = 6b 2 = 6(0.6389) 2 = 2.4492 Putting this together gives the equation f ( x) = 2.4492(0.6389) x
Section 4.1 Exponential Functions 221 Try it Now 4. Given the two points (1, 3) and (2, 4.5) find the equation of an exponential function that passes through these two points. Example 7 Find an equation for the exponential function graphed below.
The initial value for the function is not clear in this graph, so we will instead work using two clearer points. There are three fairly clear points: (-1, 1), (1, 2), and (3, 4). As we saw in the last example, two points are sufficient to find the equation for a standard exponential, so we will use the latter two points. Substituting in (1,2) gives 2 = ab1 Substituting in (3,4) gives 4 = ab 3 Solving the first equation for a gives a =
2 . b
Substituting this expression for a into the second equation: 4 = ab 3 2 3 2b 3 4= b = b b 2 4 = 2b
Simplify the right-hand side
2 = b2 b=± 2 Since we restrict ourselves to positive values of b, we will use b = 2 . We can then go back and find a: 2 2 a= = = 2 b 2 This gives us a final equation of f ( x) = 2 ( 2 ) x .
222 Chapter 4 Compound Interest In the bank certificate of deposit (CD) example earlier in the section, we encountered compound interest. Typically bank accounts and other savings instruments in which earnings are reinvested, such as mutual funds and retirement accounts, utilize compound interest. The term compounding comes from the behavior that interest is earned not on the original value, but on the accumulated value of the account. In the example from earlier, the interest was compounded monthly, so we took the annual interest rate, usually called the nominal rate or annual percentage rate (APR) and divided by 12, the number of compounds in a year, to find the monthly interest. The exponent was then measured in months. Generalizing this, we can form a general formula for compound interest. If the APR is written in decimal form as r, and there are k compounding periods per year, then the interest per compounding period will be r/k. Likewise, if we are interested in the value after t years, then there will be kt compounding periods in that time. Compound Interest Formula Compound Interest can be calculated using the formula kt
r A(t ) = a1 + k Where A(t) is the account value t is measured in years a is the starting amount of the account, often called the principal r is the annual percentage rate (APR), also called the nominal rate k is the number of compounding periods in one year
Example 8 If you invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years? Since we are starting with $3000, a = 3000 Our interest rate is 3%, so r = 0.03 Since we are compounding quarterly, we are compounding 4 times per year, so k = 4 We want to know the value of the account in 10 years, so we are looking for A(10), the value when t = 10. 0.03 A(10) = 30001 + 4
4 (10 )
= $4045.05
The account will be worth $4045.05 in 10 years.
Section 4.1 Exponential Functions 223 Example 9 A 529 plan is a college savings plan in which a relative can invest money to pay for a child’s later college tuition, and the account grows tax free. If Lily wants to set up a 529 account for her new granddaughter, wants the account to grow to $40,000 over 18 years, and she believes the account will earn 6% compounded semi-annually (twice a year), how much will Lily need to invest in the account now? Since the account is earning 6%, r = 0.06 Since interest is compounded twice a year, k = 2 In this problem, we don’t know how much we are starting with, so we will be solving for a, the initial amount needed. We do know we want the end amount to be $40,000, so we will be looking for the value of a so that A(18) = 40,000. 0.06 40,000 = A(18) = a1 + 2 40,000 = a (2.8983) 40,000 ≈ $13,801 a= 2.8983
2 (18 )
Lily will need to invest $13,801 to have $40,000 in 18 years. Try it now 5. Recalculate example 2 from above with quarterly compounding. Because of compounding throughout the year, with compound interest the actual increase in a year is more than the annual percentage rate. If $1,000 were invested at 10%, the table below shows the value after 1 year at different compounding frequencies: Frequency Annually Semiannually Quarterly Monthly Daily
Value after 1 year $1100 $1102.50 $1103.81 $1104.71 $1105.16
If we were to compute the actual percentage increase for the daily compounding, there was an increase of $105.16 from an original amount of $1,000, for a percentage increase 105.16 of = 0.10516 = 10.516% increase. This quantity is called the annual percentage 1000 yield (APY).
224 Chapter 4 Notice that given any starting amount, the amount after 1 year would be k
r A(1) = a1 + . To find the total change, we would subtract the original amount, then k to find the percentage change we would divide that by the original amount: k r a 1 + − a k r k = 1 + − 1 a k
Annual Percentage Yield The annual percentage yield is the actual percent a quantity increases in one year. It can be calculated as APY = 1 +
k
r −1 k
Notice this is equivalent to finding the value of $1 after 1 year, and subtracting the original dollar. Example 10 Bank A offers an account paying 1.2% compounded quarterly. Bank B offers an account paying 1.1% compounded monthly. Which is offering a better rate? We can compare these rates using the annual percentage yield – the actual percent increase in a year. 4
0.012 Bank A: APY = 1 + − 1 = 0.012054 = 1.2054% 4 12
0.011 Bank B: APY = 1 + − 1 = 0.011056 = 1.1056% 12 Bank B’s monthly compounding is not enough to catch up with Bank A’s better APR. Bank A offers a better rate. A Limit to Compounding As we saw earlier, the amount we earn increases as we increase the compounding frequency. The table, though, shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to believe that although increasing the frequency of compounding will increase our result, there is an upper limit to this process.
Section 4.1 Exponential Functions 225 To see this, let us examine the value of $1 invested at 100% interest for 1 year. Frequency Annual Semiannually Quarterly Monthly Daily Hourly Once per minute Once per second
Value $2 $2.25 $2.441406 $2.613035 $2.714567 $2.718127 $2.718279 $2.718282
These values do indeed appear to be approaching an upper limit. This value ends up being so important that it gets represented by its own letter, much like how π represents a number. Euler’s Number: e k
1 e is the letter used to represent the value that 1 + approaches as k gets big. k e ≈ 2.718282
Because e is often used as the base of an exponential, most scientific and graphing calculators have a button that can calculate powers of e, usually labeled ex. Some computer software instead defines a function exp(x), where exp(x) = ex. Because e arises when the time between compounds becomes very small, e allows us to define continuous growth and allows us to define a new toolkit function, f ( x) = e x . Continuous Growth Formula Continuous Growth can be calculated using the formula f ( x) = ae rx where a is the starting amount r is the continuous growth rate This type of equation is commonly used when describing quantities that change more or less continuously, like chemical reactions, growth of large populations, and radioactive decay.
226 Chapter 4 Example 11 Radon-222 decays at a continuous rate of 17.3% per day. How much will 100mg of Radon-222 decay to in 3 days? Since we are given a continuous decay rate, we use the continuous growth formula. Since the substance is decaying, we know the growth rate will be negative: r = -0.173 f (3) = 100e −0.173(3) ≈ 59.512 mg of Radon-222 will remain. Try it Now 6. Interpret the following: S (t ) = 20e0.12t if S(t) represents the growth of a substance in grams, and time is measured in days. Continuous growth is also often applied to compound interest, allowing us to talk about continuous compounding. Example 12 If $1000 is invested in an account earning 10% compounded continuously, find the value after 1 year. Here, the continuous growth rate is 10%, so r = 0.10. We start with $1000, so a = 1000. To find the value after 1 year, f (1) = 1000e 0.10 (1) ≈ $1105.17 Notice this is a $105.17 increase for the year. As a percent increase, this is 105.17 = 0.10517 = 10.517% increase over the original $1000. 1000 Notice that this value is slightly larger than the amount generated by daily compounding in the table computed earlier. The continuous growth rate is like the nominal growth rate (or APR) – it reflects the growth rate before compounding takes effect. This is different than the annual growth rate used in the formula f ( x) = a (1 + r ) x , which is like the annual percentage yield – it reflects the actual amount the output grows in a year. While the continuous growth rate in the example above was 10%, the actual annual yield was 10.517%. This means we could write two different looking but equivalent formulas for this account’s growth: using the 10% continuous growth rate f (t ) = 1000e0.10t t using the 10.517% actual annual yield rate. f (t ) = 1000(1.10517)
Section 4.1 Exponential Functions 227 Important Topics of this Section Percent growth Exponential functions Finding formulas Interpreting equations Graphs Exponential Growth & Decay Compound interest Annual Percent Yield Continuous Growth Try it Now Answers 1. A & C are exponential functions, they grow by a % not a constant number. 2. B(t) is growing faster, but after 3 years A(t) still has a higher account balance 3. 1000(0.97) 30 = 401.0071
4. f ( x) = 2(1.5) 5. $1024.25 6. An initial substance weighing 20g is growing at a continuous rate of 12% per day. x
228 Chapter 4
Section 4.1 Exercises For each table below, could the table represent a function that is linear, exponential, or neither? 1.
x 1 2 3 4 f(x) 70 40 10 -20
2.
x 1 2 3 4 g(x) 40 32 26 22
3.
1 2 3 4 x h(x) 70 49 34.3 24.01
4.
x 1 2 3 4 k(x) 90 80 70 60
5.
1 2 3 4 x m(x) 80 61 42.9 25.61
6.
1 2 3 4 x n(x) 90 81 72.9 65.61
7. A population numbers 11,000 organisms initially and grows by 8.5% each year. Write an exponential model for the population. 8. A population is currently 6,000 and has been increasing by 1.2% each day. Write an exponential model for the population. 9. The fox population in a certain region has an annual growth rate of 9 percent per year. It is estimated that the population in the year 2010 was 23,900. Estimate the fox population in the year 2018. 10. The amount of area covered by blackberry bushes in a park has been growing by 12% each year. It is estimated that the area covered in 2009 was 4,500 square feet. Estimate the area that will be covered in 2020. 11. A vehicle purchased for $32,500 depreciates at a constant rate of 5% each year. Determine the approximate value of the vehicle 12 years after purchase. 12. A business purchases $125,000 of office furniture which depreciates at a constant rate of 12% each year. Find the residual value of the furniture 6 years after purchase.
Section 4.1 Exponential Functions 229 Find a formula for an exponential function passing through the two points. 13. ( 0, 6 ) , (3, 750) 14. ( 0, 3) , (2, 75) 15. ( 0, 2000 ) , (2, 20)
16. ( 0, 9000 ) , (3, 72)
3 17. −1, , ( 3, 24 ) 2
2 18. −1, , 1,10 ( ) 5
19. ( −2, 6 ) , 3,1 ( )
20. ( −3, 4 ) , (3, 2)
21. ( 3,1 ) , (5, 4)
22. ( 2,5 ) , (6, 9)
23. A radioactive substance decays exponentially. A scientist begins with 100 milligrams of a radioactive substance. After 35 hours, 50 mg of the substance remains. How many milligrams will remain after 54 hours? 24. A radioactive substance decays exponentially. A scientist begins with 110 milligrams of a radioactive substance. After 31 hours, 55 mg of the substance remains. How many milligrams will remain after 42 hours? 25. A house was valued at $110,000 in the year 1985. The value appreciated to $145,000 by the year 2005. What was the annual growth rate between 1985 and 2005? Assume that the house value continues to grow by the same percentage. What did the value equal in the year 2010? 26. An investment was valued at $11,000 in the year 1995. The value appreciated to $14,000 by the year 2008. What was the annual growth rate between 1995 and 2008? Assume that the value continues to grow by the same percentage. What did the value equal in the year 2012? 27. A car was valued at $38,000 in the year 2003. The value depreciated to $11,000 by the year 2009. Assume that the car value continues to drop by the same percentage. What will the value be in the year 2013? 28. A car was valued at $24,000 in the year 2006. The value depreciated to $20,000 by the year 2009. Assume that the car value continues to drop by the same percentage. What will the value be in the year 2014? 29. If $4,000 is invested in a bank account at an interest rate of 7 per cent per year, find the amount in the bank after 9 years if interest is compounded annually, quarterly, monthly, and continuously.
230 Chapter 4 30. If $6,000 is invested in a bank account at an interest rate of 9 per cent per year, find the amount in the bank after 5 years if interest is compounded annually, quarterly, monthly, and continuously. 31. Find the annual percentage yield (APY) for a savings account with annual percentage rate of 3% compounded quarterly. 32. Find the annual percentage yield (APY) for a savings account with annual percentage rate of 5% compounded monthly. 33. A population of bacteria is growing according to the equation P(t ) =1 600e0.21 t , with t measured in years. Estimate when the population will exceed 7569. 34. A population of bacteria is growing according to the equation P(t ) =1 200e0.17 t , with t measured in years. Estimate when the population will exceed 3443. 35. In 1968, the U.S. minimum wage was $1.60 per hour. In 1976, the minimum wage was $2.30 per hour. Assume the minimum wage grows according to an exponential model w(t ) , where t represents the time in years after 1960. [UW] a. Find a formula for w(t ) . b. What does the model predict for the minimum wage in 1960? c. If the minimum wage was $5.15 in 1996, is this above, below or equal to what the model predicts? 36. In 1989, research scientists published a model for predicting the cumulative number 3
t − 1980 of AIDS cases (in thousands) reported in the United States: a ( t ) = 155 , 10 where t is the year. This paper was considered a “relief”, since there was a fear the correct model would be of exponential type. Pick two data points predicted by the research model a (t ) to construct a new exponential model b(t ) for the number of cumulative AIDS cases. Discuss how the two models differ and explain the use of the word “relief.” [UW]
Section 4.1 Exponential Functions 231 37. You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number of dimes. On the first square you place one dime. On the second square you stack 2 dimes. Then you continue, always doubling the number from the previous square. [UW] a. How many dimes will you have stacked on the 10th square? b. How many dimes will you have stacked on the nth square? c. How many dimes will you have stacked on the 64th square? d. Assuming a dime is 1 mm thick, how high will this last pile be? e. The distance from the earth to the sun is approximately 150 million km. Relate the height of the last pile of dimes to this distance.
232 Chapter 4
Section 4.2 Graphs of Exponential Functions Like with linear functions, the graph of an exponential function is determined by the values for the parameters in the function’s formula. To get a sense for the behavior of exponentials, let us begin by looking more closely at the function f ( x) = 2 x . Listing a table of values for this function: x -3 -2 -1 0 1 2 3 1 1 1 f(x) 1 2 4 8 8 4 2 Notice that: 1) This function is positive for all values of x. 2) As x increases, the function grows faster and faster (the rate of change increases). 3) As x decreases, the function values grow smaller, approaching zero. 4) This is an example of exponential growth. 1 Looking at the function g ( x) = 2 x -3 -2 -1
g(x)
8
4
x
2
0 1
1 1 2
2 1 4
3 1 8
Note this function is also positive for all values of x, but in this case grows as x decreases, and decreases towards zero as x increases. This is an example of exponential decay. You may notice from the table that this function appears to be the horizontal reflection of the f ( x) = 2 x table. This is in fact the case: x
f (− x) = 2
−x
1 = (2 ) = = g ( x) 2 −1 x
Looking at the graphs also confirms this relationship:
Section 4.2 Graphs of Exponential Functions 233 Consider a function for the form f ( x) = ab x . Since a, which we called the initial value in the last section, is the function value at an input of zero, a will give us the vertical intercept of the graph. From the graphs above, we can see that an exponential graph will have a horizontal asymptote on one side of the graph, and can either increase or decrease, depending upon the growth factor. This horizontal asymptote will also help us determine the long run behavior and is easy to determine from the graph. The graph will grow when the growth rate is positive, which will make the growth factor b larger than one. When it’s negative, the growth factor will be less than one. Graphical Features of Exponential Functions Graphically, in the function f ( x) = ab x a is the vertical intercept of the graph b determines the rate at which the graph grows the function will increase if b > 1 the function will decrease if 0 < b < 1 The graph will have a horizontal asymptote at y = 0 The graph will be concave up if a > 0; concave down if a < 0. The domain of the function is all real numbers The range of the function is (0, ∞) When sketching the graph of an exponential function, it can be helpful to remember that the graph will pass through the points (0, a) and (1, ab). The value b will determine the function’s long run behavior: If b > 1, as x → ∞ , f (x) → ∞ and as x → −∞ , f ( x) → 0 . If 0 < b < 1, as x → ∞ , f ( x) → 0 and as x → −∞ , f (x) → ∞ . Example 1 1 Sketch a graph of f ( x) = 4 3
x
This graph will have a vertical intercept at (0,4), and 4 pass through the point 1, . Since b < 1, the graph 3 will be decreasing towards zero. Since a > 0, the graph will be concave up. We can also see from the graph the long run behavior: as x → ∞ , f ( x) → 0 and as x → −∞ , f (x) → ∞ .
234 Chapter 4 To get a better feeling for the effect of a and b on the graph, examine the sets of graphs below. The first set shows various graphs, where a remains the same and we only change the value for b.
( 13 ) ( 12)
x
x
3x
2x 1.5 x
0.9 x
Notice that the closer the value of b is to 1, the less steep the graph will be. In the next set of graphs, a is altered and our value for b remains the same. 4 (1.2 x )
3 (1.2 x ) 2 (1.2 x ) 1.2 x
0.5 (1.2 x )
Notice that changing the value for a changes the vertical intercept. Since a is multiplying the bx term, a acts as a vertical stretch factor, not as a shift. Notice also that the long run behavior for all of these functions is the same because the growth factor did not change and none of these a values introduced a vertical flip.
Section 4.2 Graphs of Exponential Functions 235 Example 2 Match each equation with its graph. f ( x) = 2(1.3) x g ( x) = 2(1.8) x h( x) = 4(1.3) x k ( x) = 4(0.7) x
The graph of k(x) is the easiest to identify, since it is the only equation with a growth factor less than one, which will produce a decreasing graph. The graph of h(x) can be identified as the only growing exponential function with a vertical intercept at (0,4). The graphs of f(x) and g(x) both have a vertical intercept at (0,2), but since g(x) has a larger growth factor, we can identify it as the graph increasing faster. g(x) f(x) h(x)
k(x)
Try it Now 1. Graph the following functions on the same axis: f ( x) = (2) x ; g ( x) = 2(2) x ; h( x) = 2(1 / 2) x . Transformations of Exponential Graphs While exponential functions can be transformed following the same rules as any function, there are a few interesting features of transformations that can be identified. The first was seen at the beginning of the section – that a horizontal reflection is equivalent to a change in the growth factor. Likewise, since a is itself a stretch factor, a vertical stretch of an exponential corresponds with a change in the initial value of the function.
236 Chapter 4 Next consider the effect of a horizontal shift on an exponential function. Shifting the function f ( x) = 3(2) x four units to the left would give f ( x + 4) = 3(2) x + 4 . Employing exponent rules, we could rewrite this: f ( x + 4) = 3(2) x + 4 = 3(2) x (2 4 ) = 48(2) x Interestingly, it turns out that a horizontal shift of an exponential function corresponds with a change in initial value of the function. Lastly, consider the effect of a vertical shift on an exponential function. Shifting f ( x) = 3(2) x down 4 units would give the equation f ( x) = 3(2) x − 4 , yielding the graph
Notice that this graph is substantially different than the basic exponential graph. Unlike a basic exponential, this graph does not have a horizontal asymptote at y = 0; due to the vertical shift, the horizontal asymptote has also shifted to y = -4. We can see that as x → ∞ , f ( x) → ∞ and as x → −∞ , f ( x) → −4 . We have determined that a vertical shift is the only transformation of an exponential function that changes the graph in a way that cannot be achieved by altering the parameters a and b in the basic exponential function f ( x) = ab x . Transformations of Exponentials Any transformed exponential can be written in the form f ( x) = ab x + c where y = c is the horizontal asymptote. Note that, due to the shift, the vertical intercept is shifted to (0, a+c). Try it Now 2. Write the equation and graph the exponential function described as follows: f ( x) = e x is vertically stretched by a factor of 2, flipped across the y axis and shifted up 4 units.
Section 4.2 Graphs of Exponential Functions 237 Example 3 x
1 Sketch a graph of f ( x) = −3 + 4 . 2
Notice that in this exponential function, the negative in the stretch factor -3 will cause a vertical reflection, and the vertical shift up 4 will move the horizontal asymptote to x
1 y = 4. Sketching this as a transformation of g ( x) = , 2 1 The basic g ( x) = 2
x
Vertically reflected and stretched by 3
Vertically shifted up four units
Notice that while the domain of this function is unchanged, due to the reflection and shift, the range of this function is ( −∞, 4 ) . As x → ∞ , f ( x) → 4 and as x → −∞ , f ( x) → −∞ . Functions leading to graphs like the one above are common as models for learning and models of growth approaching a limit.
238 Chapter 4 Example 4 Find an equation for the graph sketched below.
Looking at this graph, it appears to have a horizontal asymptote at y = 5, suggesting an equation of the form f ( x) = ab x + 5 . To find values for a and b, we can identify two other points on the graph. It appears the graph passes through (0,2) and (-1,3), so we can use those points. Substituting in (0,2) allows us to solve for a 2 = ab 0 + 5 2 = a+5 a = −3 Substituting in (-1,3) allows us to solve for b 3 = −3b −1 + 5 −3 −2= b − 2b = −3 3 b = = 1.5 2 The final formula for our function is f ( x) = −3(1.5) x + 5 . Try it Now 3. Given the graph of the transformed exponential function, find a formula and describe the long run behavior.
Section 4.2 Graphs of Exponential Functions 239 Important Topics of this Section Graphs of exponential functions Intercept Growth factor Exponential Growth Exponential Decay Horizontal intercepts Long run behavior Transformations Try it Now Answers 1 h( x ) = 2 2
x
f ( x) = 2 x
g ( x) = 2 ( 2 x )
1.
2. f ( x) = −2e x + 4 ; 3. f ( x) = 3(.5) − x − 1 or f= ( x) 3(2 x ) − 1 ; As x → ∞ , f (x) → ∞ and as x → −∞ , f ( x) → −1
240 Chapter 4
Section 4.2 Exercises Match each function with one of the graphs below. x 1. f ( x ) = 2 ( 0.69 ) 2.
f ( x ) = 2 (1.28 )
x
3.
f ( x ) = 2 ( 0.81)
x
4.
f ( x ) = 4 (1.28 )
x
5.
f ( x ) = 2 (1.59 )
x
6.
f ( x ) = 4 ( 0.69 )
B
C
D
E
A
F
x
If all the graphs to the right have equations with form f ( x ) = ab x , 7. Which graph has the largest value for b?
B A
8. Which graph has the smallest value for b?
C
D E F
9. Which graph has the largest value for a? 10. Which graph has the smallest value for a?
Sketch a graph of each of the following transformations of f ( x ) = 2 x 11. f ( x ) = 2− x
12. g ( x ) = −2 x
13. h ( x= ) 2x + 3
14. f ( x= ) 2x − 4
15. f ( x ) = 2 x − 2
16. k ( x ) = 2 x −3
Starting with the graph of f ( x ) = 4 x , find a formula for the function that results from 17. Shifting f ( x) 4 units upwards 18. Shifting f ( x) 3 units downwards 19. Shifting f ( x) 2 units left 20. Shifting f ( x) 5 units right 21. Reflecting f ( x) about the x-axis 22. Reflecting f ( x) about the y-axis
Section 4.2 Graphs of Exponential Functions 241 Describe the long run behavior, as x → ∞ and x → −∞ of each function 23. f ( x ) = 24. f ( x ) = −5 ( 4 x ) − 1 −2 ( 3x ) + 2 x
x
1 25. = f ( x) 3 − 2 2
1 26. = f ( x) 4 +1 4
27. = f ( x ) 3( 4) + 2
28. f ( x ) = −2 ( 3) − 1
−x
−x
Find a formula for each function graphed as a transformation of f ( x ) = 2 x .
29.
30.
31.
32.
Find an equation for the exponential function graphed.
33.
34.
35.
36.
242 Chapter 4
Section 4.3 Logarithmic Functions A population of 50 flies is expected to double every week, leading to a function of the form f ( x) = 50(2) x , where x represents the number of weeks that have passed. When will this population reach 500? Trying to solve this problem leads to: Dividing both sides by 50 to isolate the exponential 500 = 50(2) x 10 = 2 x
While we have set up exponential models and used them to make predictions, you may have noticed that solving exponential equations has not yet been mentioned. The reason is simple: none of the algebraic tools discussed so far are sufficient to solve exponential equations. Consider the equation 2 x = 10 above. We know that 2 3 = 8 and 2 4 = 16 , so it is clear that x must be some value between 3 and 4 since g ( x) = 2 x is increasing. We could use technology to create a table of values or graph to better estimate the solution.
From the graph, we could better estimate the solution to be around 3.3. This result is still fairly unsatisfactory, and since the exponential function is one-to-one, it would be great to have an inverse function. None of the functions we have already discussed would serve as an inverse function and so we must introduce a new function, named log as the inverse of an exponential function. Since exponential functions have different bases, we will define corresponding logarithms of different bases as well. Logarithm The logarithm (base b) function, written log b ( x ) , is the inverse of the exponential function (base b), b x . Since the logarithm and exponential are inverses, it follows that: Properties of Logs: Inverse Properties log b (b x ) = x b logb x = x
Section 4.3 Logarithmic Functions 243 Recall also from the definition of an inverse function that if f (a ) = c , then f Applying this to the exponential and logarithmic functions:
−1
(c ) = a .
Logarithm Equivalent to an Exponential The statement b a = c is equivalent to the statement log b (c) = a . Alternatively, we could show this by starting with the exponential function c = b a , then taking the log base b of both sides, giving log b (c) = log b b a . Using the inverse property of logs we see that log b (c) = a . Since log is a function, it is most correctly written as log b (c) , using parentheses to denote function evaluation, just as we would with f(c). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written as log b c . Example 1 Write these exponential equations as logarithmic equations: 23 = 8
5 2 = 25
23 = 8
is equivalent to log 2 (8) = 3
5 2 = 25
is equivalent to log 5 (25) = 2
10−4 =
1 10000
1 is equivalent to log10 = −4 10000
Example 2 Write these logarithmic equations as exponential equations: 1 log 3 (9) = 2 log 6 6 = 2
( )
log 6
( 6 ) = 12
log 3 (9) = 2
is equivalent to 61 / 2 = 6 is equivalent to 3 2 = 9
Try it Now Write the exponential equation 4 2 = 16 as a logarithmic equation.
10 − 4 =
1 10000
244 Chapter 4 By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic and exponential equations by rewriting. Example 3 Solve log 4 ( x ) = 2 for x. By rewriting this expression as an exponential, 4 2 = x , so x = 16 Example 4 Solve 2 x = 10 for x. By rewriting this expression as a logarithm, we get x = log 2 (10) While this does define a solution, and an exact solution at that, you may find it somewhat unsatisfying since it is difficult to compare this expression to the decimal estimate we made earlier. Also, giving an exact expression for a solution is not always useful – often we really need a decimal approximation to the solution. Luckily, this is a task calculators and computers are quite adept at. Unluckily for us, most calculators and computers will only evaluate logarithms of two bases. Happily, this ends up not being a problem, as we’ll see briefly. Common and Natural Logarithms The common log is the logarithm with base 10, and is typically written log(x) . The natural log is the logarithm with base e, and is typically written ln(x) . Example 5 Evaluate log(1000) using the definition of the common log. To evaluate log(1000) , we can say x = log(1000) , then rewrite into exponential form using the common log base of 10. 10 x = 1000 From this, we might recognize that 1000 is the cube of 10, so x = 3. We also can use the inverse property of logs to write log10 (10 3 ) = 3
Values of the common log number number as log(number) exponential 1000 103 3 2 100 10 2 10 101 1 0 1 10 0 -1 0.1 10 -1 0.01 10-2 -2 -3 0.001 10 -3
Section 4.3 Logarithmic Functions 245 Try it Now 2. Evaluate log(1000000) . Example 6 Evaluate ln e .
( )
( )
( )
We can rewrite ln e as ln e1 / 2 . Since ln is a log base e, we can use the inverse 1 property for logs: ln (e1 / 2 ) = log e (e1 / 2 ) = . 2 Example 7 Evaluate log(500) using your calculator or computer. Using a computer, we can evaluate log(500) ≈ 2.69897 To utilize the common or natural logarithm functions to evaluate expressions like log 2 (10) , we need to establish some additional properties. Properties of Logs: Exponent Property log b (A r ) = r log b ( A) To show why this is true, we offer a proof. Since the logarithmic and exponential functions are inverses, b logb A = A .
(
So A r = b logb A
)
r
Utilizing the exponential rule that states ( x p ) = x pq , q
(
)
r
A r = b logb A = b r logb A So then log b (A r ) = log b b r logb A Again utilizing the inverse property on the right side yields the result log b (A r ) = r log b A
(
)
Example 8 Rewrite log 3 (25) using the exponent property for logs. Since 25 = 52, log 3 (25) = log 3 (5 2 ) = 2 log 3 5
246 Chapter 4 Example 9 Rewrite 4 ln( x) using the exponent property for logs.
( )
Using the property in reverse, 4 ln( x) = ln x 4 Try it Now
1 3. Rewrite using the exponent property for logs: ln 2 x
.
The exponent property allows us to find a method for changing the base of a logarithmic expression. Properties of Logs: Change of Base log c ( A) log b ( A) = log c (b) Proof: Let log b ( A) = x . Rewriting as an exponential gives b x = A . Taking the log base c of both sides of this equation gives log c b x = log c A Now utilizing the exponent property for logs on the left side, x log c b = log c A Dividing, we obtain log c A log c A or replacing our expression for x, log b A = x= log c b log c b With this change of base formula, we can finally find a good decimal approximation to our question from the beginning of the section. Example 10 Evaluate log 2 (10) using the change of base formula. According to the change of base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e: log e 10 ln 10 = log 2 10 = log e 2 ln 2 Using our calculators to evaluate this,
Section 4.3 Logarithmic Functions 247 ln 10 2.30259 ≈ ≈ 3.3219 ln 2 0.69315
This finally allows us to answer our original question – the population of flies we discussed at the beginning of the section will take 3.32 weeks to grow to 500. Example 11 Evaluate log 5 (100) using the change of base formula. We can rewrite this expression using any other base. If our calculators are able to evaluate the common logarithm, we could rewrite using the common log, base 10.
log 5 (100) =
log10 100 2 ≈ = 2.861 log10 5 0.69897
While we were able to solve the basic exponential equation 2 x = 10 by rewriting in logarithmic form and then using the change of base formula to evaluate the logarithm, the proof of the change of base formula illuminates an alternative approach to solving exponential equations. Solving exponential equations: 1. Isolate the exponential expressions when possible 2. Take the logarithm of both sides 3. Utilize the exponent property for logarithms to pull the variable out of the exponent 4. Use algebra to solve for the variable. Example 12 Solve 2 x = 10 for x. Using this alternative approach, rather than rewrite this exponential into logarithmic form, we will take the logarithm of both sides of the equation. Since we often wish to evaluate the result to a decimal answer, we will usually utilize either the common log or natural log. For this example, we’ll use the natural log: Utilizing the exponent property for logs, ln 2 x = ln(10) Now dividing by ln(2), x ln(2 ) = ln(10) ln(10) = x ≈ 2.861 ln ( 2 )
( )
Notice that this result matches the result we found using the change of base formula.
248 Chapter 4 Example 13 In the first section, we predicted the population (in billions) of India t years after 2008 by using the function f (t ) = 1.14(1 + 0.0134) t . If the population continues following this trend, when will the population reach 2 billion? We need to solve for the t so that f(t) = 2 Divide by 1.14 to isolate the exponential expression 2 = 1.14(1.0134) t 2 Take the logarithm of both sides of the equation = 1.0134 t 1.14 2 t Apply the exponent property on the right side ln = ln (1.0134 ) 1 . 14 2 Divide both sides by ln(1.0134) ln = t ln(1.0134 ) 1.14 2 ln 1.14 t= ≈ 42.23 years ln(1.0134 ) If this growth rate continues, the model predicts the population of India will reach 2 billion about 42 years after 2008, or approximately in the year 2050. Try it Now 4. Solve 5(0.93) x = 10 . In addition to solving exponential equations, logarithmic expressions are common in many physical situations. Example 14 In chemistry, pH is a measure of the acidity or basicity of a liquid. The pH is related to the concentration of hydrogen ions, [H+], measured in moles per liter, by the equation pH = − log H + .
(
)
If a liquid has concentration of 0.0001 moles per liber, determine the pH. Determine the hydrogen ion concentration of a liquid with pH of 7. To answer the first question, we evaluate the expression − log(0.0001) . While we could use our calculators for this, we do not really need them here, since we can use the inverse property of logs: − log(0.0001) = − log 10 −4 = −(−4) = 4
(
)
Section 4.3 Logarithmic Functions 249
(
)
To answer the second question, we need to solve the equation 7 = − log H + . Begin by isolating the logarithm on one side of the equation by multiplying both sides by -1: −7 = log H +
(
)
Rewriting into exponential form yields the answer + −7 H= 10 = 0.0000001 moles per liter. Logarithms also provide us a mechanism for finding continuous growth models for exponential growth given two data points. Example 15 A population grows from 100 to 130 in 2 weeks. Find the continuous growth rate. Measuring t in weeks, we are looking for an equation P(t ) = ae rt so that P(0) = 100 and P(2) = 130. Using the first pair of values, 100 = ae r ⋅0 , so a = 100. Using the second pair of values, Divide by 100 130 = 100e r ⋅2 130 Take the natural log of both sides = er2 100 Use the inverse property of logs ln(1.3) = ln e r 2 ln(1.3) = 2r ln(1.3) r= ≈ 0.1312 2 This population is growing at a continuous rate of 13.12% per week.
( )
In general, we can relate the standard form of an exponential with the continuous growth form by noting (using k to represent the continuous growth rate to avoid the confusion of using r in two different ways in the same formula): a (1 + r ) x = ae kx (1 + r ) x = e kx 1 + r = ek Using this, we see that it is always possible to convert from the continuous growth form of an exponential to the standard form and vice versa. Remember that the continuous growth rate k represents the nominal growth rate before accounting for the effects of continuous compounding, while r represents the actual percent increase in one time unit (one week, one year, etc.).
250 Chapter 4 Example 16 A company’s sales can be modeled by the function S (t ) = 5000e 0.12t , with t measured in years. Find the annual growth rate. Noting that 1 + r = e k , then r = e 0.12 − 1 = 0.1275 , so the annual growth rate is 12.75%. The sales function could also be written in the form S (t ) = 5000(1 + 0.1275) t . Important Topics of this Section The Logarithmic function as the inverse of the exponential function Writing logarithmic & exponential expressions Properties of logs Inverse properties Exponential properties Change of base Common log Natural log Solving exponential equations Try it Now Answers 1. log 4 (16) = 2 = log 4 4 2 = 2 log 4 4 2. 6 3. − 2 ln( x) ln(2) 4. ≈ −9.5513 ln(0.93)
Section 4.3 Logarithmic Functions 251
Section 4.3 Exercises Rewrite each equation in exponential form 1. log 4 (q ) = m 2. log 3 (t ) = k 3. log a (b) = c
5. log ( v ) = t
6. log ( r ) = s
4. log p ( z ) = u
7. ln ( w ) = n
8. ln ( x ) = y
Rewrite each equation in logarithmic form. 9. 4 x = y 10. 5 y = x 11. c d = k
12. n z = L
13. 10a = b
14. 10 p = v
15. e k = h
16. e y = x
Solve for x. 17. log 3 ( x ) = 2
18. log 4 ( x) = 3
19. log 2 ( x) = −3
20. log 5 ( x) = −1
21. log ( x ) = 3
22. log ( x ) = 5
23. ln ( x ) = 2
24. ln ( x ) = −2
Simplify each expression using logarithm properties. 25. log 5 ( 25 ) 29. log 6
26. log 2 ( 8 )
( 6)
30. log 5
33. log ( 0.001)
( 5) 3
34. log ( 0.00001)
Evaluate using your calculator. 38. log (1045 ) 37. log ( 0.04 )
1 27. log 3 27
1 28. log 6 36
31. log (10, 000 )
32. log (100 )
35. ln ( e −2 )
36. ln ( e3 )
39. ln (15 )
40. ln ( 0.02 )
Solve each equation for the variable.
1 15
41. 5 x = 14
42. 3x = 23
43. 7 x =
45. e5 x = 17
46. e3 x = 12
47. 34 x−5 = 38
44. 3x =
48. 42 x−3 = 44
49. 1000 (1.03) = 5000
50. 200 (1.06 ) = 550
51. 3 (1.04 ) = 8
52. 2 (1.08 ) = 7
53. 50e −0.12t = 10
54. 10e −0.03t = 4
t
3t
x
1 55. 10 − 8 = 5 2
1 4
t
4t
x
1 56. 100 − 100 = 70 4
252 Chapter 4 Convert the equation into continuous growth form, f ( t ) = ae kt . 57. f ( t ) = 300 ( 0.91) 59. f ( t ) = 10 (1.04 )
t
t
58. f ( t ) = 120 ( 0.07 )
t
60. f ( t ) = 1400 (1.12 )
t
Convert the equation into annual growth form, f ( t ) = abt . 61. f ( t ) =1 50e0.06t
62. f ( t ) = 100e0.12t
63. f ( t ) = 50e −0.012t
64. f ( t ) = 80e −0.85t
65. The population of Kenya was 39.8 million in 2009 and has been growing by about 2.6% each year. If this trend continues, when will the population exceed 45 million? 66. The population of Algeria was 34.9 million in 2009 and has been growing by about 1.5% each year. If this trend continues, when will the population exceed 45 million? 67. The population of Seattle grew from 563,374 in 2000 to 608,660 in 2010. If the population continues to grow exponentially at the same rate, when will the population exceed 1 million people? 68. The median household income (adjusted for inflation) in Seattle grew from $42,948 in 1990 to $45,736 in 2000. If it continues to grow exponentially at the same rate, when will median income exceed $50,000? 69. A scientist begins with 100 mg of a radioactive substance. After 4 hours, it has decayed to 80 mg. How long after the process began will it take to decay to 15 mg? 70. A scientist begins with 100 mg of a radioactive substance. After 6 days, it has decayed to 60 mg. How long after the process began will it take to decay to 10 mg? 71. If $1000 is invested in an account earning 3% compounded monthly, how long will it take the account to grow in value to $1500? 72. If $1000 is invested in an account earning 2% compounded quarterly, how long will it take the account to grow in value to $1300?
Section 4.4 Logarithmic Properties 253
Section 4.4 Logarithmic Properties In the previous section, we derived two important properties of logarithms, which allowed us to solve some basic exponential and logarithmic equations. Properties of Logs Inverse Properties: log b (b x ) = x b logb x = x
Exponential Property: log b A r = r log b ( A)
( )
Change of Base: log c ( A) log b ( A) = log c (b) While these properties allow us to solve a large number of problems, they are not sufficient to solve all problems involving exponential and logarithmic equations. Properties of Logs Sum of Logs Property: log b ( A) + log b (C ) = log b ( AC ) Difference of Logs Property: A log b ( A) − log b (C ) = log b C It’s just as important to know what properties logarithms do not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(A + B) ≠ log(A) + log(B). To help in this process we offer a proof to help solidify our new rules and show how they follow from properties you’ve already seen. Let a = log b ( A) and c = log b (C ) , so by definition of the logarithm, b a = A and b c = C
254 Chapter 4 Using these expressions, AC = b a b c Using exponent rules on the right, AC = b a + c Taking the log of both sides, and utilizing the inverse property of logs, log b ( AC ) = log b (b a + c ) = a + c Replacing a and c with their definition establishes the result log b ( AC ) = log b A + log b C The proof for the difference property is very similar. With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs. Example 1 Write log 3 (5) + log 3 (8) − log 3 (2 ) as a single logarithm. Using the sum of logs property on the first two terms, log 3 (5) + log 3 (8) = log 3 (5 ⋅ 8) = log 3 (40) This reduces our original expression to log 3 (40) − log 3 (2 ) Then using the difference of logs property, 40 log 3 (40 ) − log 3 (2 ) = log 3 = log 3 (20 ) 2 Example 2 Evaluate 2 log(5) + log(4 ) without a calculator by first rewriting as a single logarithm. On the first term, we can use the exponent property of logs to write 2 log(5) = log 5 2 = log(25)
( )
With the expression reduced to a sum of two logs, log(25) + log(4 ) , we can utilize the sum of logs property log(25) + log(4 ) = log(4 ⋅ 25) = log(100) Since 100 = 102, we can evaluate this log without a calculator: log(100) = log 10 2 = 2
( )
Try it Now 1. Without a calculator evaluate by first rewriting as a single logarithm: log 2 (8) + log 2 (4 )
Section 4.4 Logarithmic Properties 255 Example 3
x4 y as a sum or difference of logs Rewrite ln 7 First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write x4 y = ln x 4 y − ln(7) ln 7
( )
Then seeing the product in the first term, we use the sum property ln x 4 y − ln(7) = ln x 4 + ln( y ) − ln(7)
( )
( )
Finally, we could use the exponent property on the first term ln x 4 + ln( y ) − ln(7) = 4 ln( x) + ln( y ) − ln(7)
( )
Interestingly, solving exponential equations was not the reason logarithms were originally developed. Historically, up until the advent of calculators and computers, the power of logarithms was that these log properties reduced multiplication, division, roots, or powers to be evaluated using addition, subtraction, division and multiplication, respectively, which are much easier to compute without a calculator. Large books were published listing the logarithms of numbers, such as in the table to the right. To find the product of two numbers, the sum of log property was used. Suppose for example we didn’t know the value of 2 times 3. Using the sum property of logs: log(2 ⋅ 3) = log(2) + log(3)
value 1 2 3 4 5 6 7 8 9 10
log(value) 0.0000000 0.3010300 0.4771213 0.6020600 0.6989700 0.7781513 0.8450980 0.9030900 0.9542425 1.0000000
Using the log table, log(2 ⋅ 3) = log(2) + log(3) = 0.3010300 + 0.4771213 = 0.7781513 We can then use the table again in reverse, looking for 0.7781513 as an output of the logarithm. From that we can determine: log(2 ⋅ 3) = 0.7781513 = log(6) . By doing addition and the table of logs, we were able to determine 2 ⋅ 3 = 6 . Likewise, to compute a cube root like 3 8 1 1 log(3 8 ) == log(81 / 3 ) = log(8) = (0.9030900) = 0.3010300 = log(2) 3 3 3 So 8 = 2 .
256 Chapter 4 Although these calculations are simple and insignificant they illustrate the same idea that was used for hundreds of years as an efficient way to calculate the product, quotient, roots, and powers of large and complicated numbers, either using tables of logarithms or mechanical tools called slide rules. These properties still have other practical applications for interpreting changes in exponential and logarithmic relationships. Example 4 Recall that in chemistry, pH = − log H + . If the concentration of hydrogen ions in a liquid is doubled, what is the affect on pH?
(
)
Suppose C is the original concentration of hydrogen ions, and P is the original pH of the liquid, so P = − log(C ) . If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is pH = − log(2C ) Using the sum property of logs, pH = − log(2C ) = −(log(2) + log(C ) ) = − log(2) − log(C ) Since P = − log(C ) , the new pH is pH = P − log(2) = P − 0.301 When the concentration of hydrogen ions is doubled, the pH decreases by 0.301. Log properties in solving equations The logarithm properties often arise when solving problems involving logarithms. Example 5 Solve log(50 x + 25) − log( x) = 2 . In order to rewrite in exponential form, we need a single logarithmic expression on the left side of the equation. Using the difference property of logs, we can rewrite the left side: 50 x + 25 log =2 x Rewriting in exponential form reduces this to an algebraic equation: 50 x + 25 = 10 2 = 100 x
Section 4.4 Logarithmic Properties 257 Solving, 50 x + 25 = 100 x 25 = 50 x 25 1 x= = 50 2 Checking this answer in the original equation, we can verify there are no domain issues, and this answer is correct. Try it Now 2. Solve log( x 2 − 4) = 1 + log( x + 2) . More complex exponential equations can often be solved in more than one way. In the following example, we will solve the same problem in two ways – one using logarithm properties, and the other using exponential properties. Example 6a In 2008, the population of Kenya was approximately 38.8 million, and was growing by 2.64% each year, while the population of Sudan was approximately 41.3 million and growing by 2.24% each year 2. If these trends continue, when will the population of Kenya match that of Sudan? We start by writing an equation for each population in terms of t, the number of years after 2008. Kenya = (t ) 38.8(1 + 0.0264)t
Sudan = (t ) 41.3(1 + 0.0224)t To find when the populations will be equal, we can set the equations equal 38.8(1.0264)t = 41.3(1.0224)t For our first approach, we take the log of both sides of the equation log ( 38.8(1.0264)t ) = log ( 41.3(1.0224)t ) Utilizing the sum property of logs, we can rewrite each side, log(38.8) + log (1.0264t ) = log(41.3) + log (1.0224t ) Then utilizing the exponent property, we can pull the variables out of the exponent 2
World Bank, World Development Indicators, as reported on http://www.google.com/publicdata, retrieved August 24, 2010
258 Chapter 4
log(38.8) + t log (1.0264 ) = log(41.3) + t log (1.0224 ) Moving all the terms involving t to one side of the equation and the rest of the terms to the other side, t log (1.0264 ) − t log (1.0224 ) = log(41.3) − log(38.8) Factoring out the t on the left, t ( log (1.0264 ) − log (1.0224 ) ) = log(41.3) − log(38.8) Dividing to solve for t log(41.3) − log(38.8) t ≈ 15.991 years until the populations will be equal. log (1.0264 ) − log (1.0224 ) Example 6b Solve the problem above by rewriting before taking the log. Starting at the equation 38.8(1.0264)t = 41.3(1.0224)t Divide to move the exponential terms to one side of the equation and the constants to the other side 1.0264t 41.3 = 1.0224t 38.8 Using exponent rules to group on the left, t
1.0264 41.3 = 1.0224 38.8
Taking the log of both sides 1.0264 t 41.3 log = log 1.0224 38.8 Utilizing the exponent property on the left, 1.0264 41.3 t log = log 1.0224 38.8 Dividing gives 41.3 log 38.8 ≈ 15.991 years = t 1.0264 log 1.0224
Section 4.4 Logarithmic Properties 259 While the answer does not immediately appear identical to that produced using the previous method, note that by using the difference property of logs, the answer could be rewritten: 41.3 log log(41.3) − log(38.8) 38.8 = t = 1.0264 log(1.0264) − log(1.0224) log 1.0224 While both methods work equally well, it often requires fewer steps to utilize algebra before taking logs, rather than relying solely on log properties. Try it Now 3. Tank A contains 10 liters of water, and 35% of the water evaporates each week. Tank B contains 30 liters of water, and 50% of the water evaporates each week. In how many weeks will the tanks contain the same amount of water? Important Topics of this Section Inverse Exponential Change of base Sum of logs property Difference of logs property Solving equations using log rules Try it Now Answers 1. 5 2. 12 3. 4.1874 weeks
260 Chapter 4
Section 4.4 Exercises Simplify to a single logarithm, using logarithm properties. 1. log 3 ( 28 ) − log 3 ( 7 ) 2. log 3 ( 32 ) − log 3 ( 4 ) 1 3. − log 3 7
1 4. − log 4 5
1 5. log 3 + log 3 ( 50 ) 10
6. log 4 ( 3) + log 4 (7)
7.
1 log 7 ( 8 ) 3
8.
1 log 5 ( 36 ) 2
9. log ( 2 x 4 ) + log ( 3 x5 )
10. ln ( 4 x 2 ) + ln ( 3 x3 )
11. ln ( 6 x9 ) − ln ( 3 x 2 )
12. log (12 x 4 ) − log ( 4 x )
13. 2 log ( x ) + 3log ( x + 1)
14. 3log ( x ) + 2 log ( x 2 )
1 15. log ( x ) − log ( y ) + 3log ( z ) 2
1 16. 2 log ( x ) + log ( y ) − log ( z ) 3
Use logarithm properties to expand each expression. x15 y13 a 2b3 18. log 5 17. log 19 c z
a −2b3 20. ln −5 c
a −2 19. ln −4 5 b c x 3 y −4
)
22. log
y 23. ln y 1− y
x 24. ln 2 1− x
21. log
( (
25. log x 2 y 3 3 x 2 y 5
)
( (
x −3 y 2
)
26. log x3 y 4 7 x3 y 9
)
Section 4.4 Logarithmic Properties 261 Solve each equation for the variable. 27. 44 x −7 = 39 x −6
28. 22 x −5 = 73 x −7
29. 17 (1.14 ) = 19 (1.16 )
30. 20 (1.07 ) = 8 (1.13)
x
x
x
31. 5e0.12t = 10e0.08t
32. 3e0.09t = e0.14t
33. log 2 ( 7 x + 6 ) = 3
34. log 3 (2 x + 4) = 2
35. 2 ln ( 3x ) + 3 = 1
36. 4 ln ( 5 x ) + 5 = 2
( )
x
( )
37. log x3 = 2
38. log x5 = 3
39. log ( x ) + log ( x + 3) = 3
40. log ( x + 4 ) + log ( x ) = 9
1 41. log ( x + 4 ) − log ( x + 3) =
2 42. log ( x + 5 ) − log ( x + 2 ) =
43. log 6 ( x 2 ) − log 6 ( x + 1) = 1
44. log 3 ( x 2 ) − log 3 ( x + 2) = 5
45. log ( x + 12= ) log ( x ) + log (12 )
46. log ( x + 15= ) log ( x ) + log (15)
47. ln ( x ) + ln ( x − 3) = ln ( 7 x )
48. ln ( x ) + ln ( x − 6 ) = ln ( 6 x )
262 Chapter 4
Section 4.5 Graphs of Logarithmic Functions Recall that the exponential function f ( x) = 2 x produces this table of values x f(x)
-3 1 8
-2 1 4
-1 1 2
0
1
2
3
1
2
4
8
Since the logarithmic function is an inverse of the exponential, g ( x) = log 2 ( x) produces the table of values 1 1 1 x 1 2 4 8 8 4 2 g(x) -3 -2 -1 0 1 2 3 In this second table, notice that 1) As the input increases, the output increases. 2) As input increases, the output increases more slowly. 3) Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is (0, ∞) . 4) Since the exponential function can accept all real numbers as inputs, the logarithm can output any real number, so the range is all real numbers or (−∞, ∞) . Sketching the graph, notice that as the input approaches zero from the right, the output of the function grows very large in the negative direction, indicating a vertical asymptote at x = 0. In symbolic notation we write as x → 0 + , f ( x) → −∞ , and as x → ∞, f ( x) → ∞
Graphical Features of the Logarithm Graphically, in the function g ( x) = log b ( x) The graph has a horizontal intercept at (1, 0) The graph has a vertical asymptote at x = 0 The graph is increasing and concave down The domain of the function is x > 0, or (0, ∞) The range of the function is all real numbers, or (−∞, ∞) When sketching a general logarithm with base b, it can be helpful to remember that the graph will pass through the points (1, 0) and (b, 1).
Section 4.5 Graphs of Logarithmic Functions 263 To get a feeling for how the base affects the shape of the graph, examine the graphs below. log 2 ( x) ln( x) log( x)
Notice that the larger the base, the slower the graph grows. For example, the common log graph, while it grows without bound, it does so very slowly. For example, to reach an output of 8, the input must be 100,000,000. Another important observation made was the domain of the logarithm. Like the reciprocal and square root functions, the logarithm has a restricted domain which must be considered when finding the domain of a composition involving a log. Example 1 Find the domain of the function f ( x) = log(5 − 2 x) The logarithm is only defined with the input is positive, so this function will only be defined when 5 − 2 x > 0 . Solving this inequality, − 2 x > −5 5 x< 2 The domain of this function is x <
5 5 , or in interval notation, − ∞, 2 2
Try it Now 1. Find the domain of the function f ( x) = log( x − 5) + 2 ; before solving this as an inequality, consider how the function has been transformed.
264 Chapter 4 Transformations of the Logarithmic Function Transformations can be applied to a logarithmic function using the basic transformation techniques, but as with exponential functions, several transformations result in interesting relationships.
log c x 1 = log c x log c b log c b From this, we can see that log b x is a vertical stretch or compression of the graph of the log c x graph. This tells us that a vertical stretch or compression is equivalent to a change of base. For this reason, we typically represent all graphs of logarithmic functions in terms of the common or natural log functions. First recall the change of base property tells us that log b x =
Next, consider the effect of a horizontal compression on the graph of a logarithmic function. Considering f ( x) = log(cx) , we can use the sum property to see f ( x) = log(cx) = log(c) + log( x) Since log(c) is a constant, the effect of a horizontal compression is the same as the effect of a vertical shift. Example 2 Sketch f ( x) = ln( x) and g ( x) = ln( x) + 2 . Graphing these, g ( x) = ln( x) + 2 f ( x) = ln( x)
Note that, this vertical shift could also be written as a horizontal compression: g ( x) = ln( x) + 2 = ln( x) + ln(e 2 ) = ln(e 2 x) . While a horizontal stretch or compression can be written as a vertical shift, a horizontal reflection is unique and separate from vertical shifting. Finally, we will consider the effect of a horizontal shift on the graph of a logarithm
Section 4.5 Graphs of Logarithmic Functions 265 Example 3 Sketch a graph of f ( x) = ln( x + 2) . This is a horizontal shift to the left by 2 units. Notice that none of our logarithm rules allow us rewrite this in another form, so the effect of this transformation is unique. Shifting the graph,
Notice that due to the horizontal shift, the vertical asymptote shifted to x = -2, and the domain shifted to (−2, ∞) . Combining these transformations, Example 4 Sketch a graph of f ( x) = 5 log(− x + 2) . Factoring the inside as f ( x) = 5 log(−( x − 2)) reveals that this graph is that of the common logarithm, horizontally reflected, vertically stretched by a factor of 5, and shifted to the right by 2 units. The vertical asymptote will be shifted to x = 2, and the graph will have domain (∞, 2) . A rough sketch can be created by using the vertical asymptote along with a couple points on the graph, such as f (1) = 5 log(−1 + 2) = 5 log(1) = 0 f (−8) = 5 log(−(−8) + 2) = 5 log(10) = 5
Try it Now 2. Sketch a graph of the function f ( x) = −3 log( x − 2) + 1 .
266 Chapter 4 Transformations of Logs Any transformed logarithmic function can be written in the form ) a log ( − ( x − b ) ) + k if horizontally reflected, f= ( x) a log( x − b) + k , or f ( x= where x = b is the vertical asymptote. Example 5 Find an equation for the logarithmic function graphed below.
This graph has a vertical asymptote at x = –2 and has been vertically reflected. We do not know yet the vertical shift (equivalent to horizontal stretch) or the vertical stretch (equivalent to a change of base). We know so far that the equation will have form f ( x) = −a log( x + 2) + k It appears the graph passes through the points (–1, 1) and (2, –1). Substituting in (–1, 1), 1 = − a log(−1 + 2) + k 1 = − a log(1) + k 1= k Next, substituting in (2, –1), − 1 = −a log(2 + 2) + 1 − 2 = −a log(4) a=
2 log(4)
2 log( x + 2) + 1 . log(4) This could also be written as f ( x) = −2 log 4 ( x + 2) + 1 . This gives us the equation f ( x) = −
Flashback 3. Write the domain and range of the function graphed in Example 5, and describe its long run behavior.
Section 4.5 Graphs of Logarithmic Functions 267 Important Topics of this Section Graph of the logarithmic function (domain and range) Transformation of logarithmic functions Creating graphs from equations Creating equations from graphs Try it Now Answers 1. Domain: {x| x > 5}
2. Flashback Answers 3. Domain: {x|x>-2}, Range: all real numbers; As x → −2 + , f ( x) → ∞ and as x → ∞, f ( x) → −∞ .
268 Chapter 4
Section 4.5 Exercises For each function, find the domain and the vertical asymptote. f ( x ) log ( x − 5 ) f ( x ) log ( x + 2 ) 1. = 2. = 3. f = ( x ) ln ( 3 − x )
4. f = ( x ) ln ( 5 − x )
5. = f ( x ) log ( 3 x + 1)
6. = f ( x ) log ( 2 x + 5 )
7. f ( x= ) 3log ( − x ) + 2
8. f ( x= ) 2 log ( − x ) + 1
Sketch a graph of each pair of function. 9. f ( x ) log 10. f ( x ) log = = g ( x) = = g ( x) ( x ) , ln ( x) 2 ( x ), log 4 ( x) Sketch each transformation. 11. f ( x ) = 2 log ( x )
12. f ( x ) = 3ln ( x )
13. f ( x= ) ln ( − x )
14. f ( x ) = − log ( x )
f ( x ) log 2 ( x + 2) 15. =
f ( x ) log 3 ( x + 4 ) 16. =
Find a formula for the transformed logarithm graph shown.
17.
18.
19.
20.
Section 4.5 Graphs of Logarithmic Functions 269 Find a formula for the transformed logarithm graph shown.
21.
22.
23.
24.
270 Chapter 4
Section 4.6 Exponential and Logarithmic Models While we have explored some basic applications of exponential and logarithmic functions, in this section we explore some important applications in more depth. Radioactive Decay In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes change over time. One of the common terms associated with radioactive decay is halflife. Half Life The half-life of a radioactive isotope is the time it takes for half the substance to decay. Given the basic exponential growth/decay equation h(t ) = ab t , half-life can be found by 1 solving for when half the original amount remains; by solving a = a (b) t , or more 2 1 simply = b t . Notice how the initial amount is irrelevant when solving for half-life. 2 Example 1 Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210? We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant a. Then the decay of Bismuth-210 can be described by the equation Q(d ) = a (0.87) d . To find the half-life, we want to determine when the remaining quantity is half the 1 original: a . Solving, 2 1 Dividing by a, a = a (0.87) d 2 1 Take the log of both sides = 0.87 d 2 1 Use the exponent property of logs log = log 0.87 d 2 1 Divide to solve for d log = d log(0.87 ) 2
(
)
Section 4.6 Exponential and Logarithmic Models 271 1 log 2 ≈ 4.977 days = d log ( 0.87 )
This tells us that the half-life of Bismuth-210 is approximately 5 days. Example 2 Cesium-137 has a half-life of about 30 years. If you begin with 200mg of cesium-137, how much will remain after 30 years? 60 years? 90 years? Since the half-life is 30 years, after 30 years, half the original amount, 100mg, will remain. After 60 years, another 30 years have passed, so during that second 30 years, another half of the substance will decay, leaving 50mg. After 90 years, another 30 years have passed, so another half of the substance will decay, leaving 25mg. Example 3 Cesium-137 has a half-life of about 30 years. Find the annual decay rate. Since we are looking for an annual decay rate, we will use an equation of the form Q(t ) = a (1 + r ) t . We know that after 30 years, half the original amount will remain. Using this information 1 Dividing by a a = a (1 + r ) 30 2 1 Taking the 30th root of both sides = (1 + r ) 30 2 1 30 = 1+ r Subtracting one from both sides, 2
r = 30
1 − 1 ≈ −0.02284 2
This tells us cesium-137 is decaying at an annual rate of 2.284% per year. Try it Now Chlorine-36 is eliminated from the body with a biological half-life of 10 days 3. Find the daily decay rate. 3
http://www.ead.anl.gov/pub/doc/chlorine.pdf
272 Chapter 4 Example 4 Carbon-14 is a radioactive isotope that is present in organic materials, and is commonly used for dating historical artifacts. Carbon-14 has a half-life of 5730 years. If a bone fragment is found that contains 20% of its original carbon-14, how old is the bone? To find how old the bone is, we first will need to find an equation for the decay of the carbon-14. We could either use a continuous or annual decay formula, but opt to use the continuous decay formula since it is more common in scientific texts. The half life tells us that after 5730 years, half the original substance remains. Solving for the rate, 1 Dividing by a a = ae r 5730 2 1 Taking the natural log of both sides = e r 5730 2 1 Use the inverse property of logs on the right side ln = ln e r 5730 2 1 Divide by 5730 ln = 5730r 2 1 ln 2 r = ≈ −0.000121 5730
(
)
Now we know the decay will follow the equation Q(t ) = ae −0.000121t . To find how old the bone fragment is that contains 20% of the original amount, we solve for t so that Q(t) = 0.20a. 0.20a = ae −0.000121t 0.20 = e −0.000121t ln(0.20) = ln e −0.000121t ln(0.20) = −0.000121t ln(0.20) t= ≈ 13301 years − 0.000121
(
)
The bone fragment is about 13,300 years old. Try it Now 2. In Example 2, we learned that Cesium-137 has a half-life of about 30 years. If you begin with 200mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?
Section 4.6 Exponential and Logarithmic Models 273 Doubling Time For decaying quantities, we asked how long it takes for half the substance to decay. For growing quantities we might ask how long it takes for the quantity to double. Doubling Time The doubling time of a growing quantity is the time it takes for the quantity to double. Given the basic exponential growth equation h(t ) = ab t , doubling time can be found by solving for when the original quantity has doubled; by solving 2a = a (b) x , or more simply 2 = b x . Again notice how the initial amount is irrelevant when solving for doubling time. Example 5 Cancer cells sometimes increase exponentially. If a cancerous growth contained 300 cells last month and 360 cells this month, how long will it take for the number of cancer cells to double? Defining t to be time in months, with t = 0 corresponding to this month, we are given two pieces of data: this month, (0, 360), and last month, (-1, 300). From this data, we can find an equation for the growth. Using the form C (t ) = ab t , we know immediately a = 360, giving C (t ) = 360b t . Substituting in (-1, 300),
300 = 360b −1 360 b 360 b= = 1.2 300 This gives us the equation C (t ) = 360(1.2) t 300 =
To find the doubling time, we look for the time until we have twice the original amount, so when C(t) = 2a. 2a = a (1.2) t 2 = (1.2) t log(2 ) = log 1.2 t log(2 ) = t log(1.2 ) log(2 ) t= ≈ 3.802 months. log(1.2 ) It will take about 3.8 months for the number of cancer cells to double.
( )
274 Chapter 4
Example 6 Use of a new social networking website has been growing exponentially, with the number of new members doubling every 5 months. If the site currently has 120,000 users and this trend continues, how many users will the site have in 1 year? We can use the doubling time to find a function that models the number of site users, and then use that equation to answer the question. While we could use an arbitrary a as we have before for the initial amount, in this case, we know the initial amount was 120,000. If we use a continuous growth equation, it would look like N (t ) = 120e rt , measured in thousands of users after t months. Based on the doubling time, there would be 240 thousand users after 5 months. This allows us to solve for the continuous growth rate: 240 = 120e r 5 2 = er5 ln 2 = 5r ln 2 r= ≈ 0.1386 5 Now that we have an equation, N (t ) = 120e 0.1386t , we can predict the number of users after 12 months: N (12) = 120e 0.1386 (12 ) = 633.140 thousand users. So after 1 year, we would expect the site to have around 633,140 users. Try it Now 3. If tuition at a college is increasing by 6.6% each year, how many years will it take for tuition to double? Newton’s Law of Cooling When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off towards the surrounding air temperature. This "leveling off" will correspond to a horizontal asymptote in the graph of the temperature function. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function.
Section 4.6 Exponential and Logarithmic Models 275 Newton’s Law of Cooling The temperature of an object, T, in surrounding air with temperature Ts will behave according to the formula T (t ) = ae kt + Ts Where t is time a is a constant determined by the initial temperature of the object k is a constant, the continuous rate of cooling of the object While an equation of the form T (t ) = ab t + Ts could be used, the continuous growth form is more common. Example 7 A cheesecake is taken out of the oven with an ideal internal temperature of 165 degrees Fahrenheit, and is placed into a 35 degree refrigerator. After 10 minutes, the cheesecake has cooled to 150 degrees. If you must wait until the cheesecake has cooled to 70 degrees before you eat it, how long will you have to wait? Since the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially towards 35, following the equation T (t ) = ae kt + 35 We know the initial temperature was 165, so T (0) = 165 . Substituting in these values, 165 = ae k 0 + 35 165 = a + 35 a = 130 We were given another pair of data, T (10) = 150 , which we can use to solve for k 150 = 130e k10 + 35 115 = 130e k10 115 = e10 k 130 115 ln = 10k 130 115 ln 130 = −0.0123 k= 10 Together this gives us the equation for cooling: T (t ) = 130e −0.0123t + 35 .
276 Chapter 4 Now we can solve for the time it will take for the temperature to cool to 70 degrees. 70 = 130e −0.0123t + 35 35 = 130e −0.0123t 35 = e −0.0123t 130 35 ln = −0.0123t 130 35 ln 130 ≈ 106.68 t= − 0.0123 It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool. Of course, if you like your cheesecake served chilled, you’d have to wait a bit longer. Try it Now 4. A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees? Logarithmic Scales For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly. Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune
Distance (millions of km) 58 108 150 228 779 1430 2880 4500
Placed on a linear scale – one with equally spaced values – these values get bunched up. Mercury Venus Earth Mars 0
500
Jupiter
1000
1500
Neptune
Uranus
Saturn
2000
2500
3000
3500
4000
4500 distance
Section 4.6 Exponential and Logarithmic Models 277 However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent. 4 Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune
Distance (millions of km) 58 108 150 228 779 1430 2880 4500
Mercury Venus
1.5
1.75
Earth
2
2.25
Mars
2.5
log(distance) 1.76 2.03 2.18 2.36 2.89 3.16 3.46 3.65
Jupiter Saturn Uranus Neptune 2.75
3
3.25
3.5
3.75
104=10000
103=100
102=100
4 log(distance)
Sometimes, as shown above, the scale on a logarithmic number line will show the log values, but more commonly the original values are listed as powers of 10, as shown below. P 10-2
10-1
100
A
B
101
102
103
104
C
D
105
106
107
Example 8 Estimate the value of point P on the log scale above The point P appears to be half way between -2 and -1 in log value, so if V is the value of this point, Rewriting in exponential form, log(V ) ≈ −1.5 V ≈ 10 −1.5 = 0.0316
4
It is interesting to note the large gap between Mars and Jupiter on the log number line. The asteroid belt, which scientists believe consists of the remnants of an ancient planet, is located there.
278 Chapter 4 Example 9 Place the number 6000 on a logarithmic scale. Since log(6000) ≈ 3.8 , this point would belong on the log scale about here: 6000 10-2
10-1
100
101
102
103
104
105
106
107
Try it Now 5. Plot the data in the table below on a logarithmic scale 5. Approximate Sound Pressure Source of Sound/Noise in µPa (micro Pascals) Launching of the Space Shuttle 2,000,000,000 Full Symphony Orchestra 2,000,000 Diesel Freight Train at High Speed at 25 m 200,000 Normal Conversation 20,000 Soft Whispering at 2 m in Library 2,000 Unoccupied Broadcast Studio 200 Softest Sound a human can hear 20
Notice that on the log scale above Example 8, the visual distance on the scale between points A and B and between C and D is the same. When looking at the values these points correspond to, notice B is ten times the value of A, and D is ten times the value of C. A visual linear difference between points corresponds to a relative (ratio) change between the corresponding values. Logarithms are useful for showing these relative changes. For example, comparing $1,000,000 to $10,000, the first is 100 times larger than the second. 1,000,000 = 100 = 10 2 10,000 Likewise, comparing $1000 to $10, the first is 100 times larger than the second. 1,000 = 100 = 10 2 10 When one quantity is roughly ten times larger than another, we say it is one order of magnitude larger. In both cases described above, the first number was two orders of magnitude larger than the second. 5
From http://www.epd.gov.hk/epd/noise_education/web/ENG_EPD_HTML/m1/intro_5.html, retrieved Oct 2, 2010
Section 4.6 Exponential and Logarithmic Models 279 Notice that the order of magnitude can be found as the common logarithm of the ratio of the quantities. On the log scale above, B is one order of magnitude larger than A, and D is one order of magnitude larger than C. Orders of Magnitude Given two values A and B, to determine how many orders of magnitude A is greater than B, A Difference in orders of magnitude = log B Example 10 On the log scale above Example 8, how many orders of magnitude larger is C than B? The value B corresponds to 10 2 = 100 The value C corresponds to 10 5 = 100,000 100,000 10 5 = 1000 = 2 = 10 3 . The log of this value is 3. 100 10 C is three orders of magnitude greater than B, which can be seen on the log scale by the visual difference between the points on the scale. The relative change is
Try it Now 6. Using the table from Try it Now #5, what is the difference of order of magnitude between the softest sound a human can hear and the launching of the space shuttle? An example of a logarithmic scale is the Moment Magnitude Scale (MMS) used for earthquakes. This scale is commonly and mistakenly called the Richter Scale, which was a very similar scale succeeded by the MMS. Moment Magnitude Scale For an earthquake with seismic moment S, a measurement of earth movement, the MMS value, or magnitude of the earthquake, is 2 S M = log 3 S0 Where S 0 = 1016 is a baseline measure for the seismic moment.
280 Chapter 4 Example 11 If one earthquake has a MMS magnitude of 6.0, and another has a magnitude of 8.0, how much more powerful (in terms of earth movement) is the second earthquake? Since the first earthquake has magnitude 6.0, we can find the amount of earth movement. The value of S0 is not particularly relevant, so we will not replace it with its value. 2 S 6.0 = log 3 S0
S 3 6.0 = log 2 S0 S 9 = log S0 S = 10 9 S0
S = 10 9 S 0
Doing the same with the second earthquake with a magnitude of 8.0, 2 S 8.0 = log 3 S0 S = 1012 S 0
From this, we can see that this second value’s earth movement is 1000 times as large as the first earthquake. Example 12 One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth movement as the first earthquake, find the magnitude of the second quake. Since the first quake has magnitude 3.0,
Section 4.6 Exponential and Logarithmic Models 281
3.0 =
2 S log 3 S 0
S 3 3.0 = log 2 S0 S 4.5 = log S0 S 10 4.5 = S0
S = 10 4.5 S 0 Since the second earthquake has twice as much earth movement, for the second quake, S = 2 ⋅ 10 4.5 S 0 Finding the magnitude, 2 2 ⋅ 10 4.5 S 0 M = log 3 S0 M =
2 log 2 ⋅ 10 4.5 ≈ 3.201 3
(
)
The second earthquake with twice as much earth movement will have a magnitude of about 3.2. In fact, using log properties, we could show that whenever the earth movement doubles, the magnitude will increase by about 0.201: 2 2S 2 S M = log = log 2 ⋅ 3 S0 3 S0 2 2 S M = log(2) + log 3 3 S0 M =
S 2 log( 2 ) log + 3 S0
2 S M = 0.201 + log 3 S0
This illustrates the most important feature of a log scale: that multiplying the quantity being considered will add to the scale value, and vice versa.
282 Chapter 4 Important Topics of this Section Radioactive decay Half life Doubling time Newton’s law of cooling Logarithmic Scales Orders of Magnitude Moment Magnitude scale Try it Now Answers 1 1. r = 10 − 1 ≈ −0.067 or 6.7% is the daily rate of decay. 2 2. Less than 230 years, 229.3157 to be exact 3. It will take 10.845 years, or approximately 11 years, for tuition to double. 4. 6.026 hours 5. Broadcast Conversation Space Softest room Soft Symphony Train Shuttle Sound Whisper 101
102
103
104
105
106
107
108
109
1010
2 x10 9 = 10 8 The sound pressure in µPa created by launching the space shuttle is 8 1 2 x10 orders of magnitude greater than the sound pressure in µPa created by the softest sound a human ear can hear. 6.
Section 4.6 Exponential and Logarithmic Models 283
Section 4.6 Exercises 1. You go to the doctor and he injects you with 13 milligrams of radioactive dye. After 12 minutes, 4.75 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm whenever more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived and the amount of dye decays exponentially? 2. You take 200 milligrams of a headache medicine, and after 4 hours, 120 milligrams remain in your system. If the effects of the medicine wear off when less than 80 milligrams remain, when will you need to take a second dose, assuming the amount of medicine in your system decays exponentially? 3. The half-life of Radium-226 is 1590 years. If a sample initially contains 200 mg, how many milligrams will remain after 1000 years? 4. The half-life of Fermium-253 is 3 days. If a sample initially contains 100 mg, how many milligrams will remain after 1 week? 5. The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample still contains 2 mg. What was the initial mass of the sample, and how much will remain after another 3 days? 6. The half-life of Nobelium-259 is 58 minutes. After 3 hours a sample still contains10 mg. What was the initial mass of the sample, and how much will remain after another 8 hours? 7. A scientist begins with 250 grams of a radioactive substance. After 225 minutes, the sample has decayed to 32 grams. Find the half-life of this substance. 8. A scientist begins with 20 grams of a radioactive substance. After 7 days, the sample has decayed to 17 grams. Find the half-life of this substance. 9. A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.)
284 Chapter 4 10. A wooden artifact from an archeological dig contains 15 percent of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.) 11. A bacteria culture initially contains 1500 bacteria and doubles in size every half hour. Find the size of the population after: a) 2 hours b) 100 minutes 12. A bacteria culture initially contains 2000 bacteria and doubles in size every half hour. Find the size of the population after: a) 3 hours b) 80 minutes 13. The count of bacteria in a culture was 800 after 10 minutes and 1800 after 40 minutes. a. What was the initial size of the culture? b. Find the doubling time. c. Find the population after 105 minutes. d. When will the population reach 11000? 14. The count of bacteria in a culture was 600 after 20 minutes and 2000 after 35 minutes. a. What was the initial size of the culture? b. Find the doubling time. c. Find the population after 170 minutes. d. When will the population reach 12000? 15. Find the time required for an investment to double in value if invested in an account paying 3% compounded quarterly. 16. Find the time required for an investment to double in value if invested in an account paying 4% compounded monthly 17. The number of crystals that have formed after t hours is given by n ( t ) = 20e0.013t . How long does it take the number of crystals to double? 18. The number of building permits in Pasco t years after 1992 roughly followed the equation n ( t ) = 400e0.143t . What is the doubling time?
Section 4.6 Exponential and Logarithmic Models 285 19. A turkey is pulled from the oven when the internal temperature is 165° Fahrenheit, and is allowed to cool in a 75° room. If the temperature of the turkey is 145° after half an hour, a. What will the temperature be after 50 minutes? b. How long will it take the turkey to cool to 110°? 20. A cup of coffee is poured at 190° Fahrenheit, and is allowed to cool in a 70° room. If the temperature of the coffee is 170° after half an hour, a. What will the temperature be after 70 minutes? b. How long will it take the coffee to cool to 120°? 21. The population of fish in a farm-stocked lake after t years could be modeled by the 1000 equation P ( t ) = . 1 + 9e −0.6t a. Sketch a graph of this equation. b. What is the initial population of fish? c. What will the population be after 2 years? d. How long will it take for the population to reach 900? 22. The number of people in a town who have heard a rumor after t days can be modeled 500 by the equation N ( t ) = . 1 + 49e −0.7 t a. Sketch a graph of this equation. b. How many people started the rumor? c. How many people have heard the rumor after 3 days? d. How long will it take until 300 people have heard the rumor? Find the value of the number shown on each logarithmic scale 23.
24.
25.
26.
Plot each set of approximate values on a logarithmic scale. 27. Intensity of sounds: Whisper: 10−10 / W m 2 , Vacuum: 10−4 W / m 2 , Jet: 102 / W m2 28. Mass: Amoeba: 10−5 g , Human: 105 g , Statue of Liberty: 108 g
286 Chapter 4 29. The 1906 San Francisco earthquake had a magnitude of 7.9 on the MMS scale. Later there was an earthquake with magnitude 4.7 that caused only minor damage. How many times more intense was the San Francisco earthquake than the second one? 30. The 1906 San Francisco earthquake had a magnitude of 7.9 on the MMS scale. Later there was an earthquake with magnitude 6.5 that caused less damage. How many times more intense was the San Francisco earthquake than the second one? 31. One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. 32. One earthquake has magnitude 4.8 on the MMS scale. If a second earthquake has 1200 times as much energy as the first, find the magnitude of the second quake. 33. A colony of yeast cells is estimated to contain 106 cells at time t = 0. After collecting experimental data in the lab, you decide that the total population of cells at time t hours is given by the function f ( t ) = 106 e0.495105t . [UW] a. How many cells are present after one hour? b. How long does it take of the population to double?. c. Cherie, another member of your lab, looks at your notebook and says: “That formula is wrong, my calculations predict the formula for the number of yeast cells is given by the function. f ( t ) = 106 ( 2.042727 )
0.693147 t
.” Should you be
worried by Cherie’s remark? d. Anja, a third member of your lab working with the same yeast cells, took these two measurements: 7.246 ×106 cells after 4 hours; 16.504 ×106 cells after 6 hours. Should you be worried by Anja’s results? If Anja’s measurements are correct, does your model over estimate or under estimate the number of yeast cells at time t? 34. As light from the surface penetrates water, its intensity is diminished. In the clear waters of the Caribbean, the intensity is decreased by 15 percent for every 3 meters of depth. Thus, the intensity will have the form of a general exponential function. [UW] a. If the intensity of light at the water’s surface is I 0 , find a formula for I (d ) , the intensity of light at a depth of d meters. Your formula should depend on I 0 and d. b. At what depth will the light intensity be decreased to 1% of its surface intensity?
Section 4.6 Exponential and Logarithmic Models 287 35. Myoglobin and hemoglobin are oxygen-carrying molecules in the human body. Hemoglobin is found inside red blood cells, which flow from the lungs to the muscles through the bloodstream. Myoglobin is found in muscle cells. The function p calculates the fraction of myoglobin saturated with oxygen at a = Y M = ( p) 1+ p given pressure p Torrs. For example, at a pressure of 1 Torr, M(1) = 0.5, which means half of the myoglobin (i.e. 50%) is oxygen saturated. (Note: More precisely, you need to use something called the “partial pressure”, but the distinction is not important for
p 2.8 calculates the fraction 262.8 + p 2.8 of hemoglobin saturated with oxygen at a given pressure p. [UW] a. The graphs of M ( p ) and H ( p ) are given here on the domain 0 ≤ p ≤ 100; which is which? b. If the pressure in the lungs is 100 Torrs, what is the level of oxygen saturation of the hemoglobin in the lungs? this problem.) Likewise, the function = Y H= ( p)
c. The pressure in an active muscle is 20 Torrs. What is the level of oxygen saturation of myoglobin in an active muscle? What is the level of hemoglobin in an active muscle? d. Define the efficiency of oxygen transport at a given pressure p to be M ( p ) − H ( p ) . What is the oxygen transport efficiency at 20 Torrs? At 40 Torrs? At 60 Torrs? Sketch the graph of M ( p ) − H ( p ) ; are there conditions under which transport efficiency is maximized (explain)? 36. The length of some fish are modeled by a von Bertalanffy growth function. For
Pacific halibut, this function has the form = L ( t ) 200 (1 − 0.957e −0.18t ) where L(t ) is the length (in centimeters) of a fish t years old. [UW] a. What is the length of a newborn halibut at birth? b. Use the formula to estimate the length of a 6–year–old halibut. c. At what age would you expect the halibut to be 120 cm long? d. What is the practical (physical) significance of the number 200 in the formula for L(t ) ?
288 Chapter 4 37. A cancer cell lacks normal biological growth regulation and can divide continuously. Suppose a single mouse skin cell is cancerous and its mitotic cell cycle (the time for the cell to divide once) is 20 hours. The number of cells at time t grows according to an exponential model. [UW] a. Find a formula C (t ) for the number of cancerous skin cells after t hours. b. Assume a typical mouse skin cell is spherical of radius 50×10−4 cm. Find the combined volume of all cancerous skin cells after t hours. When will the volume of cancerous cells be 1 cm3? 38. A ship embarked on a long voyage. At the start of the voyage, there were 500 ants in the cargo hold of the ship. One week into the voyage, there were 800 ants. Suppose the population of ants is an exponential function of time. [UW] a. How long did it take the population to double? b. How long did it take the population to triple? c. When were there be 10,000 ants on board? d. There also was an exponentially growing population of anteaters on board. At the start of the voyage there were 17 anteaters, and the population of anteaters doubled every 2.8 weeks. How long into the voyage were there 200 ants per anteater? 39. The populations of termites and spiders in a certain house are growing exponentially. The house contains 100 termites the day you move in. After 4 days, the house contains 200 termites. Three days after moving in, there are two times as many termites as spiders. Eight days after moving in, there were four times as many termites as spiders. How long (in days) does it take the population of spiders to triple? [UW]
Section 4.7 Fitting Exponential Models to Data 289
Section 4.7 Fitting Exponential Models to Data In the previous section, we saw number lines using logarithmic scales. It is also common to see two dimensional graphs with one or both axes using a logarithmic scale. One common use of a logarithmic scale on the vertical axis is to graph quantities that are changing exponentially, since it helps reveal relative differences. This is commonly used in stock charts, since values historically have grown exponentially over time. Both stock charts below show the Dow Jones Industrial Average, from 1928 to 2010.
Both charts have a linear horizontal scale, but the first graph has a linear vertical scale, while the second has a logarithmic vertical scale. The first scale is the one we are more familiar with, and shows what appears to be a strong exponential trend, at least up until the year 2000.
290 Chapter 4 Example 1 There were stock market drops in 1929 and 2008. Which was larger? In the first graph, the stock market drop around 2008 looks very large, and in terms of dollar values, it was indeed a large drop. However the second graph shows relative changes, and the drop in 2009 seems less major on this graph, and in fact the drop starting in 1929 was, percentage-wise, much more significant. Specifically, in 2008, the Dow value dropped from about 14,000 to 8,000, a drop of 6,000. This is obviously a large value drop, and amounts to about a 43% drop. In 1929, the Dow value dropped from a high of around 380 to a low of 42 by July of 1932. While value-wise this drop of 338 is much smaller than the 2008 drop, it corresponds to a 89% drop, a much larger relative drop than in 2008. The logarithmic scale shows these relative changes. The second graph above, in which one axis uses a linear scale and the other axis uses a logarithmic scale, is an example of a semi-log graph. Semi-log and Log-log Graphs A semi-log graph is a graph with one axis using a linear scale and one axis using a logarithmic scale. A log-log graph is a graph with both axes using logarithmic scales. Example 2 Plot 5 points on the graph of f ( x) = 3(2) x on a semi-log graph with a logarithmic scale on the vertical axis. To do this, we need to find 5 points on the graph, then calculate the logarithm of the output value. Arbitrarily choosing 5 input values, x -3 -1 0 2 5
f(x)
3 8 3 3(2) −1 = 2 0 3(2) = 3 3(2) −3 =
log(f(x)) -0.426 0.176 0.477
2
1.079
5
1.982
3(2) = 12 3(2) = 96
Section 4.7 Fitting Exponential Models to Data 291 Plotting these values on a semi-log graph, log(f(x)) 3 2 1
x
0 -4
-3
-1
-2
1
0
2
4
3
5
6
-1
Notice that on this semi-log scale, values from the exponential function appear linear. We can show this behavior is expected by utilizing logarithmic properties. For the function f ( x) = ab x , finding log(f(x)) gives Utilizing the sum property of logs, log( f ( x) ) = log ab x log( f ( x) ) = log(a ) + log b x Now utilizing the exponent property, log( f ( x) ) = log(a ) + x log(b )
( )
( )
This relationship is linear, with log(a) as the vertical intercept, and log(b) as the slope. This relationship can also be utilized in reverse. Example 3 An exponential graph is plotted on a semi-log graph below. Find a formula for the exponential function g(x) that generated this graph. log(g(x)) 5 4 3 2 1 0 -4
-3
-2
-1 -1 0
1
2
3
4
x
-2 -3
The graph is linear, with vertical intercept at (0, 1). Looking at the change between the 3 points (0, 1) and (4, 4), we can determine the slope of the line is . Since the output is 4 3 log(g(x)), this leads to the equation log( g ( x) ) = 1 + x . 4
292 Chapter 4 We can solve this formula for g(x) by rewriting in exponential form and simplifying: 3 Rewriting as an exponential, log( g ( x) ) = 1 + x 4 g ( x) = 10
3 1+ x 4
g ( x) = 101 ⋅ 10
Breaking this apart using exponent rules, 3 x 4
Using exponent rules to group the second factor,
3 g ( x) = 101 ⋅ 10 4 x g ( x) = 10(5.623)
x
Evaluating the powers of 10,
Try it Now 1. An exponential graph is plotted on a semi-log graph below. Find a formula for the exponential function g(x) that generated this graph. log(g(x)) 5
4 3 2
x
1 0 -4
-3
-2
-1
-1
0
1
2
3
4
Fitting Exponential Functions to Data Some technology options provide dedicated functions for finding exponential functions that fit data, but many only provide functions for fitting linear functions to data. The semi-log scale provides us with a method to fit an exponential function to data by building upon the techniques we have for fitting linear functions to data. To fit an exponential function to a set of data using linearization 1. Find the log of the data output values 2. Find the linear equation that fits the (input, log(output)) pairs. This equation will be of the form log(f(x)) = b + mx 3. Solve this equation for the exponential function f(x)
Section 4.7 Fitting Exponential Models to Data 293 Example 4 The table below shows the cost in dollars per megabyte of storage space on computer hard drives from 1980 to 2004 6, and the data is shown on a standard graph to the right, with the input changed to years after 1980 Year Cost per MB 250 1980 192.31 200 1984 87.86 150 1988 15.98 100 1992 4 1996 0.173 50 2000 0.006849 0 2004 0.001149 0 4 8 12 16 20 24 This data appears to be decreasing exponentially. To find a function that models this decay, we would start by finding the log of the costs. 3 Year Cost per MB log(Cost) 2 1980 192.31 2.284002 1 1984 87.86 1.943791 0 1988 15.98 1.203577 0 4 8 12 16 20 24 -1 1992 4 0.60206 -2 1996 0.173 -0.76195 2000 0.006849 -2.16437 -3 2004 0.001149 -2.93952 -4 As expected, the graph of the log of costs appears fairly linear, suggesting an exponential function will fit the original data will fit reasonably well. Using technology, we can find a linear equation to fit the log(Cost) values. Using t as years after 1980, linear regression gives the equation: log(C (t )) = 2.794 − 0.231t Solving for C(t), C (t ) = 10 2.794−0.231t C (t ) = 10 2.794 ⋅ 10 −0.231t
(
C (t ) = 10 2.794 ⋅ 10 −0.231 C (t ) = 622 ⋅ (0.5877 )
)
t
t
This equation suggests that the cost per megabyte for storage on computer hard drives is decreasing by about 41% each year.
6
Selected values from http://www.swivel.com/workbooks/26190-Cost-Per-Megabyte-of-Hard-DriveSpace, retrieved Aug 26, 2010
294 Chapter 4 Using this function, we could predict the cost of storage in the future. Predicting the cost in the year 2020 (t = 40): 40 C (40) = 622 ⋅ (0.5877 ) ≈ 0.000000364 dollars per megabyte, a really small number. That is equivalent to $0.36 per terabyte of hard drive storage. Comparing the values predicted by this model to the actual data, we see the model matches the original data in order of magnitude, but the specific values appear quite different. This is, unfortunately, the best exponential model that can fit the data. It is possible that a non-exponential model would fit the data better, or there could just be wide enough variability in the data that no relatively simple model would fit the data any better. Year 1980 1984 1988 1992 1996 2000 2004
Actual Cost per MB 192.31 87.86 15.98 4 0.173 0.006849 0.001149
Cost predicted by model 622.3 74.3 8.9 1.1 0.13 0.015 0.0018
Try it Now 2. The table below shows the value V, in billions of dollars, of US imports from China t years after 2000. year 2000 2001 2002 2003 2004 2005 t 0 1 2 3 4 5 V 100 102.3 125.2 152.4 196.7 243.5 This data appears to be growing exponentially. Linearize this data and build a model to predict how many billions of dollars of imports were expected in 2011. Important Topics of this Section Semi-log graph Log-log graph Linearizing exponential functions Fitting an exponential equation to data Try it Now Answers 1. f ( x) = 100(0.3162) x 2. V (t ) = 90.545(1.2078) t . Predicting in 2011, V (11) = 722.45 billion dollars
Section 4.7 Fitting Exponential Models to Data 295
Section 4.7 Exercises Graph each function on a semi-log scale, then find a formula for the linearized function in the form log ( f ( x= ) ) mx + b . 1. f ( x ) = 4 (1.3)
2. f ( x ) = 2 (1.5 )
x
3. f ( x ) = 10 ( 0.2 )
x
x
4. f ( x ) = 30 ( 0.7 )
x
The graph below is on a semi-log scale, as indicated. Find a formula for the exponential function y ( x) .
5.
6.
7.
8.
Use regression to find an exponential function that best fits the data given. 9.
2 3 4 5 6 x 1 y 1125 1495 2310 3294 4650 6361
10.
2 3 4 5 6 x 1 y 643 829 920 1073 1330 1631
11.
2 3 4 5 6 x 1 y 555 383 307 210 158 122
296 Chapter 4 12.
2 3 4 5 6 x 1 y 699 701 695 668 683 712
13. Total expenditures (in billions of dollars) in the US for nursing home care are shown below. Use regression to find an exponential function that models the data. What does the model predict expenditures will be in 2015? Year 1990 Expenditure 53
1995 74
2000 95
2003 110
2005 121
2008 138
14. Light intensity as it passes through water decreases exponentially with depth. The data below shows the light intensity (in lumens) at various depths. Use regression to find an function that models the data. What does the model predict the intensity will be at 25 feet? Depth (ft) Lumen
3 11.5
6 8.6
9 6.7
12 5.2
15 3.8
18 2.9
15. The average price of electricity (in cents per kilowatt hour) from 1990 through 2008 is given below. Determine if a linear or exponential model better fits the data, and use the better model to predict the price of electricity in 2014. Year Cost
1990 1992 1994 1996 1998 2000 2002 2004 2006 2008 7.83 8.21 8.38 8.36 8.26 8.24 8.44 8.95 10.40 11.26
16. The average cost of a loaf of white bread from 1986 through 2008 is given below. Determine if a linear or exponential model better fits the data, and use the better model to predict the price of a loaf of bread in 2016. Year Cost
1986 0.57
1988 0.66
1990 0.70
1995 0.84
1997 0.88
2000 0.99
2002 1.03
2004 0.97
2006 1.14
2008 1.42
Chapter 5: Trigonometric Functions of Angles
In the previous chapters we have explored a variety of functions which could be combined to form a variety of shapes. In this discussion, one common shape has been missing: the circle. We already know certain things about the circle, like how to find area and circumference, and the relationship between radius and diameter, but now, in this chapter, we explore the circle and its unique features that lead us into the rich world of trigonometry. Section 5.1 Circles ...................................................................................................... 297 Section 5.2 Angles ...................................................................................................... 307 Section 5.3 Points on Circles using Sine and Cosine.................................................. 321 Section 5.4 The Other Trigonometric Functions ........................................................ 333 Section 5.5 Right Triangle Trigonometry ................................................................... 343
Section 5.1 Circles To begin, we need to remember how to find distances. Starting with the Pythagorean Theorem, which relates the sides of a right triangle, we can find the distance between two points. Pythagorean Theorem The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle will equal the square of the hypotenuse of the triangle. In graphical form, given the triangle shown, a 2 + b 2 = c2 .
c
a b
We can use the Pythagorean Theorem to find the distance between two points on a graph. Example 1 Find the distance between the points (-3, 2) and (2, 5). By plotting these points on the plane, we can then draw a right triangle with these points at each end of the hypotenuse. We can calculate horizontal width of the triangle to be 5 and the vertical height to be 3. From these we can find the distance between the points using the Pythagorean Theorem:
dist 2 = 5 2 + 3 2 = 34 dist = 34 This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
298 Chapter 5 Notice that the width of the triangle was calculated using the difference between the x (input) values of the two points, and the height of the triangle was found using the difference between the y (output) values of the two points. Generalizing this process gives us the general distance formula. Distance Formula The distance between two points ( x1 , y1 ) and ( x 2 , y 2 ) can be calculated as dist = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2
Try it Now 1. Find the distance between the points (1, 6) and (3, -5). Circles If we wanted to find an equation to represent a circle with a radius of r centered at a point (h, k), we notice that the distance between any point (x, y) on the circle and the center point is always the same: r. Noting this, we can use our distance formula to write an equation for the radius:
(x, y) r (h, k)
r = ( x − h) 2 + ( y − k ) 2
Squaring both sides of the equation gives us the standard equation for a circle. Equation of a Circle The equation of a circle centered at the point (h, k) with radius r can be written as ( x − h) 2 + ( y − k ) 2 = r 2 Notice that a circle does not pass the vertical line test. It is not possible to write y as a function of x or vice versa. Example 2 Write an equation for a circle centered at the point (-3, 2) with radius 4. Using the equation from above, h = -3, k = 2, and the radius r = 4. Using these in our formula, simplified a bit, this gives ( x − (−3)) 2 + ( y − 2) 2 = 4 2 ( x + 3) 2 + ( y − 2) 2 = 16
Section 5.1 Circles 299 Example 3 Write an equation for the circle graphed here. This circle is centered at the origin, the point (0, 0). By measuring horizontally or vertically from the center out to the circle, we can see the radius is 3. Using this information in our formula gives: simplified a bit, this gives ( x − 0) 2 + ( y − 0) 2 = 3 2 x2 + y2 = 9 Try it Now 2. Write an equation for a circle centered at (4, -2) with radius 6. Notice that, relative to a circle centered at the origin, horizontal and vertical shifts of the circle are revealed in the values of h and k, which are the coordinates for the center of the circle. Points on a Circle As noted earlier, an equation for a circle cannot be written so that y is a function of x or vice versa. To find coordinates on the circle given only the x or y value, we must solve algebraically for the unknown values. Example 4 Find the points on a circle of radius 5 centered at the origin with an x value of 3. We begin by writing an equation for the circle centered at the origin with a radius of 5. x 2 + y 2 = 25 Substituting in the desired x value of 3 gives an equation we can solve for y 3 2 + y 2 = 25 y 2 = 25 − 9 = 16 y = ± 16 = ±4 There are two points on the circle with an x value of 3: (3, 4) and (3, -4).
300 Chapter 5 Example 5 Find the x intercepts of a circle with radius 6 centered at the point (2, 4). We can start by writing an equation for the circle. ( x − 2) 2 + ( y − 4) 2 = 36 To find the x intercepts, we need to find the points where y = 0. Substituting in zero for y, we can solve for x. ( x − 2) 2 + (0 − 4) 2 = 36 ( x − 2) 2 + 16 = 36 ( x − 2) 2 = 20 x − 2 = ± 20 x = 2 ± 20 = 2 ± 2 5
(
)
(
The x intercepts of the circle are 2 + 2 5 ,0 and 2 − 2 5 ,0
)
Example 6 In a town, Main Street runs east to west, and Meridian Road runs north to south. A pizza store is located on Meridian 2 miles south of the intersection of Main and Meridian. If the store advertises that it delivers within a 3 mile radius, how much of Main Street do they deliver to? This type of question is one in which introducing a coordinate system and drawing a picture can help us solve the problem. We could either place the origin at the intersection of the two streets, or place the origin at the pizza store itself. It is often easier to work with circles centered at the origin, so we’ll place the origin at the pizza store, though either approach would work fine. Placing the origin at the pizza store, the delivery area with radius 3 miles can be described as the region inside the circle described by x 2 + y 2 = 9 . Main Street, located 2 miles north of the pizza store and running east to west, can be described by the equation y = 2. To find the portion of Main Street the store will deliver to, we first find the boundary of their delivery region by looking for where the delivery circle intersects Main Street. To find the intersection, we look for the points on the circle where y = 2. Substituting y = 2 into the circle equation lets us solve for the corresponding x values.
Section 5.1 Circles 301 x 2 + 22 = 9 x2 = 9 − 4 = 5 x = ± 5 ≈ ±2.236 This means the pizza store will deliver 2.236 miles down Main Street east of Meridian and 2.236 miles down Main Street west of Meridian. We can conclude that the pizza store delivers to a 4.472 mile segment of Main St. In addition to finding where a vertical or horizontal line intersects the circle, we can also find where an arbitrary line intersects a circle. Example 7 Find where the line f ( x) = 4 x intersects the circle ( x − 2) 2 + y 2 = 16 . Normally, to find an intersection of two functions f(x) and g(x) we would solve for the x value that would make the function equal by solving the equation f(x) = g(x). In the case of a circle, it isn’t possible to represent the equation as a function, but we can utilize the same idea. The output value of the line determines the y value: y = f ( x) = 4 x . We want the y value of the circle to equal the y value of the line, which is the output value of the function. To do this, we can substitute the expression for y from the line into the circle equation. ( x − 2) 2 + y 2 = 16 ( x − 2) 2 + (4 x) 2 = 16 x 2 − 4 x + 4 + 16 x 2 = 16 17 x 2 − 4 x + 4 = 16 17 x 2 − 4 x − 12 = 0
we replace y with the line formula: y = 4 x expand and simplify since this equation is quadratic, we arrange one side to be 0
Since this quadratic doesn’t appear to be easily factorable, we can use the quadratic formula to solve for x: − (−4) ± (−4) 2 − 4(17)(−12) 4 ± 832 , or approximately x ≈ 0.966 or -0.731 x= = 2(17) 34 From these x values we can use either equation to find the corresponding y values. Since the line equation is easier to evaluate, we might choose to use it: y = f (0.966) = 4(0.966) = 3.864 y = f (−0.731) = 4(−0.731) = −2.923 The line intersects the circle at the points (0.966, 3.864) and (-0.731, -2.923).
302 Chapter 5
Try it Now 3. A small radio transmitter broadcasts in a 50 mile radius. If you drive along a straight line from a city 60 miles north of the transmitter to a second city 70 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter? Important Topics of This Section Distance formula Equation of a Circle Finding the x coordinate of a point on the circle given the y coordinate or vice versa Finding the intersection of a circle and a line Try it Now Answers 1. 5 5 2. ( x − 4) 2 + ( y + 2) 2 = 36 2
60 502 gives x = 14 or x = 45.29, corresponding to points (14, 48)and x 2 + 60 − x = 70 (45.29, 21.18), resulting in a distance of 41.21 miles.
Section 5.1 Circles 303
Section 5.1 Exercises 1. Find the distance between the points (5,3) and (-1,-5). 2. Find the distance between the points (3,3) and (-3,-2). 3. Write an equation of the circle centered at (8 , -10) with radius 8. 4. Write an equation of the circle centered at (-9, 9) with radius 16. 5. Write an equation of the circle centered at (7, -2) that passes through (-10, 0). 6. Write an equation of the circle centered at (3, -7) that passes through (15, 13). 7. Write an equation for a circle where the points (2, 6) and (8, 10) lie along a diameter. 8. Write an equation for a circle where the points (-3, 3) and (5, 7) lie along a diameter. 9. Sketch a graph of ( x − 2 . = ) + ( y + 3) 9 2
2
10. Sketch a graph of ( x + 1) + 1 6 . ( y − 2) = 2
2
11. Find the y intercept(s) of the circle with center (2, 3) with radius 3. 12. Find the x intercept(s) of the circle with center (2, 3) with radius 4. 13. At what point in the first quadrant does the line with equation = y 2 x + 5 intersect a circle with radius 3 and center (0, 5)? 14. At what point in the first quadrant does the line with equation y= 2 x + intersect the circle with radius 6 and center (0, 2)? 15. At what point in the second quadrant does the line with equation = y 2 x + 5 intersect a circle with radius 3 and center (-2, 0)? 16. At what point in the first quadrant does the line with equation y= 2 x + intersect the circle with radius 6 and center (-1,0)? 17. A small radio transmitter broadcasts in a 53 mile radius. If you drive along a straight line from a city 70 miles north of the transmitter to a second city 74 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter? 18. A small radio transmitter broadcasts in a 44 mile radius. If you drive along a straight line from a city 56 miles south of the transmitter to a second city 53 miles west of the transmitter, during how much of the drive will you pick up a signal from the transmitter?
304 Chapter 5
19. A tunnel connecting two portions of a space station has a circular cross-section of radius 15 feet. Two walkway decks are constructed in the tunnel. Deck A is along a horizontal diameter and another parallel Deck B is 2 feet below Deck A. Because the space station is in a weightless environment, you can walk vertically upright along Deck A, or vertically upside down along Deck B. You have been assigned to paint “safety stripes” on each deck level, so that a 6 foot person can safely walk upright along either deck. Determine the width of the “safe walk zone” on each deck. [UW]
20. A crawling tractor sprinkler is located as pictured here, 100 feet south of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves north along the hose at the rate of ½ inch/second. The hose is perpendicular to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk. [UW] a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and find equations of the lines forming and find equations of the lines forming the north and south boundaries of the sidewalk. b) When will the water first strike the sidewalk? c) When will the water from the sprinkler fall completely north of the sidewalk? d) Find the total amount of time water from the sprinkler falls on the sidewalk. e) Sketch a picture of the situation after 33 minutes. Draw an accurate picture of the watered portion of the sidewalk. f) Find the area of grass watered after one hour.
Section 5.1 Circles 305 21. Erik’s disabled sailboat is floating anchored 3 miles East and 2 miles north of Kingston. A ferry leaves Kingston heading toward Edmonds at 12 mph. Edmonds is 6 miles due east of Kingston 1. After 20 minutes the ferry turns, heading due south. Ballard is 8 miles south and 1 mile west of Edmonds. Impose coordinates with Ballard as the origin. [UW]
a) Find equations for the lines along which the ferry is moving and draw in these lines. b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat. Looking down from above, as in the picture, the radar region looks like a circular disk. The boundary is the “edge” or circle around this disk, the interior is everything inside of the circle, and the exterior is everything outside of the circle. Give the mathematical description (an equation or inequality) of the boundary, interior and exterior of the radar zone. Sketch an accurate picture of the radar zone by determining where the line connecting Kingston and Edmonds would cross the radar zone. c) When does the ferry enter the radar zone? d) Where and when does the ferry exit the radar zone? e) How long does the ferry spend inside the radar zone?
1
This is not actually true, sorry.
306 Chapter 5 22. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular wheat field of radius 200 ft. The combine cuts a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north and 60 feet west of the westernmost edge of the field. [UW]
a) b) c) d)
When does Nora’s combine first start cutting the wheat? When does Nora’s combine first start cutting a swath 20 feet wide? Find the total amount of time wheat is being cut during this pass across the field. Estimate the area of the swath cut during this pass across the field.
23. The vertical cross-section of a drainage ditch is pictured to the right. Here, R indicates in each case the radius of a circle with R = 10 feet, where all of the indicated circle centers lie along a horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is flowing into the ditch so that the level above the bottom is rising at a rate of 2 inches per minute. [UW] a) When will the ditch be completely full? b) Find a piecewise defined function that models the vertical cross-section of the ditch. c) What is the width of the filled portion of the ditch after 1 hour and 18 minutes? d) When will the filled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide?
Section 5.2 Angles 307
Section 5.2 Angles Because many applications involving circles also involve a rotation of the circle, it is natural to introduce a measure for the rotation, or angle, between two rays (line segments) emanating from the center of a circle. The angle measurement you are most likely familiar with is degrees, so we’ll begin there. Measure of an Angle The measure of an angle is a measurement between two intersecting lines, line segments or rays, starting at the initial side and ending at the terminal side. It is a rotational measure not a linear measure.
terminal side angle initial side
Measuring Angles Degrees A degree is a measurement of angle. One full rotation around the circle is equal to 360 degrees, so one degree is 1/360 of a circle. An angle measured in degrees should always include the unit “degrees” after the number, or include the degree symbol °. For example, 90 degrees = 90° . Standard Position When measuring angles on a circle, unless otherwise directed, we measure angles in standard position: starting at the positive horizontal axis and with counter-clockwise rotation. Example 1 Give the degree measure of the angle shown on the circle. The vertical and horizontal lines divide the circle into quarters. Since one full rotation is 360 degrees= 360° , each quarter rotation is 360/4 = 90° or 90 degrees. Example 2 Show an angle of 30° on the circle. An angle of 30° is 1/3 of 90° , so by dividing a quarter rotation into thirds, we can sketch a line at 30° .
308 Chapter 5 Going Greek When representing angles using variables, it is traditional to use Greek letters. Here is a list of commonly encountered Greek letters.
θ theta
ϕ or φ
α
phi
alpha
β beta
γ gamma
Working with Angles in Degrees Notice that since there are 360 degrees in one rotation, an angle greater than 360 degrees would indicate more than 1 full rotation. Shown on a circle, the resulting direction in which this angle’s terminal side points would be the same as for another angle between 0 and 360 degrees. These angles would be called coterminal. Coterminal Angles After completing their full rotation based on the given angle, two angles are coterminal if they terminate in the same position, so their terminal sides coincide (point in the same direction). Example 3 Find an angle θ that is coterminal with 800° , where 0° ≤ θ < 360° Since adding or subtracting a full rotation, 360 degrees, would result in an angle with terminal side pointing in the same direction, we can find coterminal angles by adding or subtracting 360 degrees. An angle of 800 degrees is coterminal with an angle of 800360 = 440 degrees. It would also be coterminal with an angle of 440-360 = 80 degrees. The angle θ = 80° is coterminal with 800° . By finding the coterminal angle between 0 and 360 degrees, it can be easier to see which direction the terminal side of an angle points in. Try it Now 1. Find an angle α that is coterminal with 870° , where 0° ≤ α < 360° . On a number line a positive number is measured to the right and a negative number is measured in the opposite direction (to the left). Similarly a positive angle is measured counterclockwise and a negative angle is measured in the opposite direction (clockwise).
Section 5.2 Angles 309 Example 4 Show the angle − 45° on the circle and find a positive angle α that is coterminal and 0° ≤ α < 360° . Since 45 degrees is half of 90 degrees, we can start at the positive horizontal axis and measure clockwise half of a 90 degree angle.
315° -45°
Since we can find coterminal angles by adding or subtracting a full rotation of 360 degrees, we can find a positive coterminal angle here by adding 360 degrees: − 45° + 360° = 315° Try it Now 2. Find an angle β coterminal with −300° where 0° ≤ β < 360° . It can be helpful to have a familiarity with the frequently encountered angles in one rotation of a circle. It is common to encounter multiples of 30, 45, 60, and 90 degrees. These values are shown to the right. Memorizing these angles and understanding their properties will be very useful as we study the properties associated with angles
135°
120°
90°
60° 45°
150°
30°
180°
0°
210°
330°
225°
315° 240°
270°
300°
Angles in Radians While measuring angles in degrees may be familiar, doing so often complicates matters since the units of measure can get in the way of calculations. For this reason, another measure of angles is commonly used. This measure is based on the distance around a circle. Arclength Arclength is the length of an arc, s, along a circle of radius r subtended (drawn out) by an angle θ . It is the portion of the circumference between the initial and terminal sides of the angle.
r
s θ
310 Chapter 5 The length of the arc around an entire circle is called the circumference of a circle. The circumference of a circle is C = 2πr . The ratio of the circumference to the radius, produces the constant 2π . Regardless of the radius, this ratio is always the same, just as how the degree measure of an angle is independent of the radius. To elaborate on this idea, consider two circles, one with radius 2 and one with radius 3. Recall the circumference (perimeter) of a circle is C = 2πr , where r is the radius of the circle. The smaller circle then has circumference 2π (2) = 4π and the larger has circumference 2π (3) = 6π . Drawing a 45 degree angle on the two circles, we might be interested in the length of the arc of the circle that the angle indicates. In both cases, the 45 degree angle draws out an arc that is 1/8th of the full circumference, so for the smaller circle, the 1 1 arclength = (4π ) = π , and for the larger circle, the length of the 8 2 arc or arclength 1 3 = (6π ) = π . 8 4
45°
Notice what happens if we find the ratio of the arclength divided by the radius of the circle: 1 π 1 2 Smaller circle: = π 2 4 3 π 1 Larger circle: 4 = π 3 4 The ratio is the same regardless of the radius of the circle – it only depends on the angle. This property allows us to define a measure of the angle based on arclength.
Radians The radian measure of an angle is the ratio of the length of the circular arc subtended by the angle to the radius of the circle. In other words, if s is the length of an arc of a circle, and r is the radius of the circle, then s radian measure = r If the circle has radius 1, then the radian measure corresponds to the length of the arc.
2 3
Section 5.2 Angles 311
Because radian measure is the ratio of two lengths, it is a unitless measure. It is not necessary to write the label “radians” after a radian measure, and if you see an angle that is not labeled with “degrees” or the degree symbol, you should assume that it is a radian measure. Considering the most basic case, the unit circle (a circle with radius 1), we know that 1 rotation equals 360 degrees, 360° . We can also track one rotation around a circle by finding the circumference, C = 2πr , and for the unit circle C = 2π . These two different ways to rotate around a circle give us a way to convert from degrees to radians. 1 rotation = 360° = 2π radians ½ rotation = 180° = π radians ¼ rotation = 90° = π / 2 radians Example 5 Find the radian measure of one third of a full rotation. For any circle, the arclength along such a rotation would be one third of the 2πr 1 circumference, C = (2πr ) = . The radian measure would be the arclength divided 3 3 by the radius: 2π r 2π Radian measure = . = 3r 3 Converting Between Radians and Degrees 1 degree =
π
180
radians
or: to convert from degrees to radians, multiply by 1 radian =
180
π
π radians 180°
degrees
or: to convert from radians to degrees, multiply by
180° π radians
312 Chapter 5
Example 6 Convert
π 6
radians to degrees.
Since we are given a problem in radians and we want degrees, we multiply by
180°
π
Remember radians are a unitless measure, so we don’t need to write “radians.” π 180° π radians = ⋅ = 30 degrees. 6 6 π
.
Example 7 Convert 15 degrees to radians. In this example we start with degrees and want radians so we use the other conversion of radians.
π
180°
so that the degree units cancel and we are left with the unitless measure
15 degrees = 15° ⋅
Try it Now 3. Convert
π 180°
=
π 12
7π radians to degrees. 10
Just as we listed all the common angles in degrees on a circle, we should also list the corresponding radian values for the common measures of a circle corresponding to degree multiples of 30, 45, 60, and 90 degrees. As with the degree measurements, it would be advisable to commit these to memory. We can work with the radian measures of an angle the same way we work with degrees.
5π 6
3π 4
2π 3
π 2
π 3
π 4
π 6
π
0, 2π 7π 6
5π 4
4π 3
3π 2
5π 3
7π 4
11π 6
Section 5.2 Angles 313 Example 8 Find an angle β that is coterminal with
19π , where 0 ≤ β < 2π 4
When working in degrees, we found coterminal angles by adding or subtracting 360 degrees, a full rotation. Likewise, in radians, we can find coterminal angles by adding or subtracting full rotations of 2π radians. 19π 19π 8π 11π − 2π = − = 4 4 4 4 11π The angle is coterminal, but not less than 2π , so we subtract another rotation. 4 11π 11π 8π 3π − 2π = − = 4 4 4 4
The angle
3π 19π is coterminal with . 4 4
Try it Now 4. Find an angle φ that is coterminal with −
17π where 0 ≤ φ < 2π . 6
Arclength and Area of a Sector Recall that the radian measure of an angle was defined as the ratio of the arclength of a s circular arc to the radius of the circle, θ = . From this relationship, we can find r arclength along a circle given an angle. Arclength on a Circle The length of an arc, s, along a circle of radius r subtended by angle θ in radians is s = rθ Example 9 Mercury orbits the sun at a distance of approximately 36 million miles. In one Earth day, it completes 0.0114 rotation around the sun. If the orbit was perfectly circular, what distance through space would Mercury travel in one Earth day? To begin, we will need to convert the decimal rotation value to a radian measure. Since one rotation = 2π radians, 0.0114 rotation = 2π (0.0114) = 0.0716 radians.
314 Chapter 5 Combining this with the given radius of 36 million miles, we can find the arclength: = s (36)(0.0716) = 2.578 million miles travelled through space. Try it Now 5. Find the arclength along a circle of radius 10 subtended by an angle of 215 degrees. In addition to arclength, we can also use angles to find the area of a sector of a circle. A sector is a portion of a circle contained between two lines from the center, like a slice of pizza or pie. Recall that the area of a circle with radius r can be found using the formula A = π r 2 . If a sector is cut out by an angle of θ , measured in radians, then the fraction of full circle that
θ , since 2π is one full rotation. Thus, the area of the sector would 2π be this fraction of the whole area: 2 1 θ 2 θπ r Area of sector πr = = θ r2 = 2π 2 2π angle has cut out is
Area of a Sector The area of a sector of a circle with radius r subtended by an angle θ , measured in radians, is 1 Area of sector = θ r 2 2
r
θ
Example 10 An automatic lawn sprinkler sprays a distance of 20 feet while rotating 30 degrees. What is the area of the sector of grass the sprinkler waters? First we need to convert the angle measure into radians. Since 30 degrees is one of our common angles, you ideally should already know the equivalent radian measure, but if not we can convert: 30 degrees = 30 ⋅
π
π
=radians. 180 6
The area of the sector is= then Area
30°
1π 2 2 = (20) 104.72 ft 2 6
20 ft
Section 5.2 Angles 315 Try it Now 6. In central pivot irrigation, a large irrigation pipe on wheels rotates around a center point, as pictured here 2. A farmer has a central pivot system with a radius of 400 meters. If water restrictions only allow her to water 150 thousand square meters a day, what angle should she set the system to cover?
Linear and Angular Velocity When your car drives down a road, it makes sense to describe its speed in terms of miles per hour or meters per second. These are measures of speed along a line, also called linear velocity. When a point on a circle rotates, we would describe its angular velocity, or rotational speed, in radians per second, rotations per minute, or degrees per hour. Angular and Linear Velocity As a point moves along a circle of radius r, its angular velocity, ω , can be found as the angular rotation θ per unit time, t.
ω=
θ t
The linear velocity, v, of the point can be found as the distance travelled, arclength s, per unit time, t. s v= t Example 11 A water wheel completes 1 rotation every 5 seconds. Find the angular velocity in radians per second. 3 The wheel completes 1 rotation = 2π radians in 5 seconds, so the 2π angular velocity would be = ω ≈ 1.257 radians per second. 5 Combining the definitions above with the arclength equation, s = rθ , we can find a relationship between angular and linear velocities. The angular velocity equation can be solved for θ , giving θ = ωt . Substituting this into the arclength equation gives = s r= θ rωt . 2 3
http://commons.wikimedia.org/wiki/File:Pivot_otech_002.JPG CC-BY-SA http://en.wikipedia.org/wiki/File:R%C3%B6mische_S%C3%A4gem%C3%BChle.svg CC-BY
316 Chapter 5 Substituting this into the linear velocity equation gives s rωt v= = = rω t t Relationship Between Linear and Angular Velocity When the angular velocity is measured in radians per unit time, linear velocity and angular velocity are related by the equation v = rω Example 12 A bicycle has wheels 28 inches in diameter. A tachometer determines the wheels are rotating at 180 RPM (revolutions per minute). Find the speed the bicycle is travelling down the road. Here we have an angular velocity and need to find the corresponding linear velocity, since the linear speed of the outside of the tires is the speed at which the bicycle travels down the road. We begin by converting from rotations per minute to radians per minute. It can be helpful to utilize the units to make this conversion rotations 2π radians radians 180 360π ⋅ = minute rotation minute Using the formula from above along with the radius of the wheels, we can find the linear velocity radians inches = v (14 inches) 360π 5040π minute minute You may be wondering where the “radians” went in this last equation. Remember that radians are a unitless measure, so it is not necessary to include them. Finally, we may wish to convert this linear velocity into a more familiar measurement, like miles per hour. inches 1 feet 1 mile 60 minutes 5040π 14.99 miles per hour (mph). ⋅ ⋅ ⋅ = minute 12 inches 5280 feet 1 hour Try it Now 7. A satellite is rotating around the earth at 27,359 kilometers per hour at an altitude of 242 km above the earth. If the radius of the earth is 6378 kilometers, find the angular velocity of the satellite.
Section 5.2 Angles 317 Important Topics of This Section Degree measure of angle Radian measure of angle Conversion between degrees and radians Common angles in degrees and radians Coterminal angles Arclength Area of a sector Linear and angular velocity Try it Now Answers 1. α = 150° 2. β = 60° 3. 126° 7π 4. 6 215π 5. ≈ 37.525 18 6. 107.43° 7. 4.1328 radians per hour
318 Chapter 5
Section 5.2 Exercises 1. Indicate each angle on a circle: 30°, 300°, -135°, 70°,
2π 7π , 3 4
2. Indicate each angle on a circle: 30°, 315°, -135°, 80°,
7π 3π , 6 4
3. Convert the angle 180° to radians. 4. Convert the angle 30° to radians. 5. Convert the angle
5π from radians to degrees. 6
6. Convert the angle
11 π from radians to degrees. 6
7. Find the angle between 0° and 360° that is coterminal with a 685° angle. 8. Find the angle between 0° and 360° that is coterminal with a 451° angle. 9. Find the angle between 0° and 360° that is coterminal with a -1746° angle. 10. Find the angle between 0° and 360° that is coterminal with a -1400° angle. 11. Find the angle between 0 and 2π in radians that is coterminal with the angle 12. Find the angle between 0 and 2π in radians that is coterminal with the angle
26 π . 9
17 π . 3
13. Find the angle between 0 and 2π in radians that is coterminal with the angle −
3 π . 2
14. Find the angle between 0 and 2π in radians that is coterminal with the angle −
7 π . 6
15. On a circle of radius 7 miles, find the length of the arc that subtends a central angle of 5 radians. 16. On a circle of radius 6 feet, find the length of the arc that subtends a central angle of 1 radian.
Section 5.2 Angles 319 17. On a circle of radius 12 cm, find the length of the arc that subtends a central angle of 120 degrees. 18. On a circle of radius 9 miles, find the length of the arc that subtends a central angle of 800 degrees. 19. Find the distance along an arc on the surface of the Earth that subtends a central angle of 5 minutes (1 minute = 1/60 degree). The radius of the Earth is 3960 miles. 20. Find the distance along an arc on the surface of the Earth that subtends a central angle of 7 minutes (1 minute = 1/60 degree). The radius of the Earth is 3960 miles. 21. On a circle of radius 6 feet, what angle in degrees would subtend an arc of length 3 feet? 22. On a circle of radius 5 feet, what angle in degrees would subtend an arc of length 2 feet? 23. A sector of a circle has a central angle of 45°. Find the area of the sector if the radius of the circle is 6 cm. 24. A sector of a circle has a central angle of 30°. Find the area of the sector if the radius of the circle is 20 cm. 25. A truck with 32-in.-diameter wheels is traveling at 60 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make? 26. A bicycle with 24-in.-diameter wheels is traveling at 15 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make? 27. A wheel of radius 8 in. is rotating 15°/sec. What is the linear speed v, the angular speed in RPM, and the angular speed in rad/sec? 28. A wheel of radius 14 in. is rotating 0.5 rad/sec. What is the linear speed v, the angular speed in RPM, and the angular speed in deg/sec? 29. A CD has diameter of 120 millimeters. When playing audio, the angular speed varies to keep the linear speed constant where the disc is being read. When reading along the outer edge of the disc, the angular speed is about 200 RPM (revolutions per minute). Find the linear speed. 30. When being burned in a writable CD-R drive, the angular speed of a CD is often much faster than when playing audio, but the angular speed still varies to keep the linear speed constant where the disc is being written. When writing along the outer edge of the disc, the angular speed of one drive is about 4800 RPM (revolutions per minute). Find the linear speed.
320 Chapter 5
31. You are standing on the equator of the Earth (radius 3960 miles). What is your linear and angular speed? 32. The restaurant in the Space Needle in Seattle rotates at the rate of one revolution per hour. [UW] a) Through how many radians does it turn in 100 minutes? b) How long does it take the restaurant to rotate through 4 radians? c) How far does a person sitting by the window move in 100 minutes if the radius of the restaurant is 21 meters?
Section 5.3 Points on Circles Using Sine and Cosine 321
Section 5.3 Points on Circles Using Sine and Cosine
While it is convenient to describe the location of a point on a circle using an angle or a distance along the circle, relating this information to the x and y coordinates and the circle equation we explored in Section 5.1 is an important application of trigonometry. A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a distance of 20 miles at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat? In a general sense, to investigate this, we begin by drawing a circle centered at the origin with radius r, and marking the point on the circle indicated by some angle θ. This point has coordinates (x, y).
(x, y) r θ
If we drop a line segment vertically down from this point to the x axis, we would form a right triangle inside of the circle. No matter which quadrant our angle θ puts us in we can draw a triangle by dropping a perpendicular line segment to the x axis, keeping in mind that the values of x and y may be positive or negative, depending on the quadrant.
Additionally, if the angle θ puts us on an axis, we simply measure the radius as the x or y with the other value being 0, again ensuring we have appropriate signs on the coordinates based on the quadrant. Triangles obtained from different radii will all be similar triangles, meaning corresponding sides scale proportionally. While the lengths of the sides may change, the ratios of the side lengths will always remain constant for any given angle. To be able to refer to these ratios more easily, we will give them names. Since the ratios depend on the angle, we will write them as functions of the angle θ . Sine and Cosine For the point (x, y) on a circle of radius r at an angle of θ , we can define two important functions as the ratios of the sides of the corresponding triangle: y The sine function: sin(θ ) = r x The cosine function: cos(θ ) = r
(x, y) r θ x
y
322 Chapter 5 In this chapter, we will explore these functions using both circles and right triangles. In the next chapter we will take a closer look at the behavior and characteristics of the sine and cosine functions. Example 1 The point (3, 4) is on the circle of radius 5 at some angle θ. Find cos(θ ) and sin(θ ) . Knowing the radius of the circle and coordinates of the point, we can evaluate the cosine and sine functions as the ratio of the sides. x 3 y 4 cos(θ ) = = sin(θ ) = = r 5 r 5 There are a few cosine and sine values which we can determine fairly easily because the corresponding point on the circle falls on the x or y axis. Example 2 Find cos(90°) and sin(90°) On any circle, the terminal side of a 90 degree angle points straight up, so the coordinates of the corresponding point on the circle would be (0, r). Using our definitions of cosine and sine, x 0 cos(90°) = = = 0 r r y r sin(90°) = = = 1 r r
(0, r) r
90°
Try it Now 1. Find cosine and sine of the angle π . Notice that the definitions above can also be stated as: Coordinates of the Point on a Circle at a Given Angle On a circle of radius r at an angle of θ , we can find the coordinates of the point (x, y) at that angle using x = r cos(θ ) y = r sin(θ ) On a unit circle, a circle with radius 1, x = cos(θ ) and y = sin(θ ) .
Section 5.3 Points on Circles Using Sine and Cosine 323 Utilizing the basic equation for a circle centered at the origin, x 2 + y 2 = r 2 , combined with the relationships above, we can establish a new identity. substituting the relations above, x2 + y2 = r 2 2 2 2 simplifying, (r cos(θ )) + (r sin(θ )) = r 2 2 2 2 2 dividing by r 2 r (cos(θ )) + r (sin(θ )) = r or using shorthand notation (cos(θ )) 2 + (sin(θ )) 2 = 1 cos 2 (θ ) + sin 2 (θ ) = 1 Here cos 2 (θ ) is a commonly used shorthand notation for (cos(θ )) 2 . Be aware that many calculators and computers do not understand the shorthand notation. In Section 5.1 we related the Pythagorean Theorem a 2 + b 2 = c 2 to the basic equation of a circle x 2 + y 2 = r 2 , which we have now used to arrive at the Pythagorean Identity. Pythagorean Identity The Pythagorean Identity. For any angle θ, cos 2 (θ ) + sin 2 (θ ) = 1 . One use of this identity is that it helps us to find a cosine value of an angle if we know the sine value of that angle or vice versa. However, since the equation will yield two possible values, we will need to utilize additional knowledge of the angle to help us find the desired value. Example 3 If sin(θ ) =
3 and θ is in the second quadrant, find cos(θ ) . 7
Substituting the known value for sine into the Pythagorean identity, 2
3 cos (θ ) + = 1 7 9 =1 cos 2 (θ ) + 49 40 cos 2 (θ ) = 49 40 40 2 10 cos(θ ) = ± = ± = ± 49 7 7 Since the angle is in the second quadrant, we know the x value of the point would be negative, so the cosine value should also be negative. Using this additional information, 2 10 we can conclude that cos(θ ) = − . 7 2
324 Chapter 5 Values for Sine and Cosine At this point, you may have noticed that we haven’t found any cosine or sine values from angles not on an axis. To do this, we will need to utilize our knowledge of triangles. First, consider a point on a circle at an angle of 45 degrees, or
π
. (x, y) = (x, x) 4 At this angle, the x and y coordinates of the corresponding point 1 on the circle will be equal because 45 degrees divides the first y quadrant in half. Since the x and y values will be the same, the 45° sine and cosine values will also be equal. Utilizing the x Pythagorean Identity, π π since the sine and cosine are equal, we can cos 2 + sin 2 = 1 4 4 substitute sine with cosine π π add like terms cos 2 + cos 2 = 1 4 4 π divide 2 cos 2 = 1 4 π 1 since the x value is positive, we’ll keep the positive root cos 2 = 4 2 π cos = 4
1 2
often this value is written with a rationalized denominator
Remember, to rationalize the denominator we multiply by a term equivalent to 1 to get rid of the radical in the denominator. 1 2 2 2 π cos = = = 2 2 4 2 4 2 π Since the sine and cosine are equal, sin = as well. 2 4 The (x, y) coordinates for a point on a circle of radius 1 at an angle of 45 degrees are 2 2 2 , 2 .
Section 5.3 Points on Circles Using Sine and Cosine 325 Example 4 Find the coordinates of the point on a circle of radius 6 at an angle of
π 4
.
2 2 π π Using our new knowledge that sin = and cos = , along with our 2 2 4 4 relationships that stated x = r cos(θ ) and y = r sin(θ ) , we can find the coordinates of the point desired: 2 π x = 6 cos = 6 =3 2 2 4 2 π =3 2 y = 6 sin = 6 2 4 Try it Now 2. Find the coordinates of the point on a circle of radius 3 at an angle of 90° . Next, we will find the cosine and sine at an angle of 30 degrees, or
π
. To do this, we will first draw a 6 triangle inside a circle with one side at an angle of 30 degrees, and another at an angle of -30 degrees. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be 60 degrees.
r
60°
r
60°
(x, y) r 30°
y
60° y
Since all the angles are equal, the sides will all be equal as well. The vertical line has r length 2y, and since the sides are all equal we can conclude that 2y = r, or y = . Using 2 this, we can find the sine value: r r 1 1 π y sin = = 2 = ⋅ = 2 r 2 r 6 r
326 Chapter 5 Using the Pythagorean Identity, we can find the cosine value: π π cos 2 + sin 2 = 1 6 6 2
π 1 cos + = 1 6 2 π 3 cos 2 = 6 4 2
since the y value is positive, we’ll keep the positive root
3 3 π = cos = 4 2 6 The (x, y) coordinates for the point on a circle of radius 1 at an angle of 30 degrees are 3 1 2 ,2. By drawing a the triangle inside the unit circle with a 30 degree angle and reflecting it over the line y = x, we can find the cosine and sine for 60 degrees, or additional work. 1 2
y=x 1
1 2
30° 3 2
3 2 30° 1
π
3
, without any
y=x
60°
By this symmetry, we can see the coordinates of the point on the unit circle at an angle of 1 3 , giving 60 degrees will be , 2 2 π 1 cos = 3 2
and
3 π sin = 3 2
We have now found the cosine and sine values for all of the commonly encountered angles in the first quadrant of the unit circle. Angle 0 π π π π , or 30° , or 45° , or 60° , or 90° 6 4 2 3 Cosine 1 0 1 3 2 2 2 2 Sine 0 1 1 2 3 2 2 2
Section 5.3 Points on Circles Using Sine and Cosine 327 For any given angle in the first quadrant, there will be an angle in another quadrant with the same sine value, and yet another angle in yet another quadrant with the same cosine value. Since the sine value is the y coordinate on the unit circle, the other angle with the same sine will share the same y value, but have the opposite x value. Likewise, the angle with the same cosine will share the same x value, but have the opposite y value. As shown here, angle α has the same sine value as angle θ; the cosine values would be opposites. The angle β has the same cosine value as the angle; the sine values would be opposites. sin(θ ) = sin(α ) and cos(θ ) = − cos(α )
sin(θ ) = − sin( β ) and cos(θ ) = cos( β )
(x, y)
(x, y) α
r
r
θ
θ
β
It is important to notice the relationship between the angles. If, from the angle, you measured the smallest angle to the horizontal axis, all would have the same measure in absolute value. We say that all these angles have a reference angle of θ. Reference Angle An angle’s reference angle is the size of the smallest angle to the horizontal axis.
(x, y)
A reference angle is always an angle between 0 and 90 degrees, or 0 and
π
2
radians.
Angles share the same cosine and sine values as their reference angles, except for signs (positive or negative) which can be determined from the quadrant of the angle.
θ
θ θ
θ
328 Chapter 5 Example 5 Find the reference angle of 150 degrees. Use it to find cos(150°) and sin(150°) . 150 degrees is located in the second quadrant. It is 30 degrees short of the horizontal axis at 180 degrees, so the reference angle is 30 degrees. This tells us that 150 degrees has the same sine and cosine values as 30 degrees, except 3 1 for sign. We know that sin(30°) = and cos(30°) = . Since 150 degrees is in the 2 2 second quadrant, the x coordinate of the point on the circle would be negative, so the cosine value will be negative. The y coordinate is positive, so the sine value will be positive. 3 1 sin(150°) = and cos(150°) = − 2 2 − 3 1 The (x, y) coordinates for the point on a unit circle at an angle of 150° are , . 2 2 Using symmetry and reference angles, we can fill in cosine and sine values at the rest of the special angles on the unit circle. Take time to learn the (x, y) coordinates of all of the major angles in the first quadrant! 2π 1 3 120°, , − , 3 2 2 135°,
150°,
90°,
π 2
, ( 0, 1)
60°,
π 1
3π 2 2 , − , 4 2 2
3 , , 3 2 2 45°,
π
4 2
5π 3 1 , − , 6 2 2
30°,
,
2
2
π 3 1 6
,
2
,
2
0°, 0, (1, 0 )
180°, π , ( −1, 0 )
3 1 210°, , − ,− 6 2 2 5π 2 2 225°, , − ,− 4 2 2 4π 1 3 240°, , − ,− 3 2 2
2
,
360°, 2π , (1, 0 )
7π
330°,
315°,
270°,
3π 2
, ( 0, −1)
300°,
5π 3
1
,
2
,−
7π 4
6
2
,
2
3 2
11π
3
1 ,− 2 2
,
,−
2 2
Section 5.3 Points on Circles Using Sine and Cosine 329 Example 6 Find the coordinates of the point on a circle of radius 12 at an angle of
7π . 6
Note that this angle is in the third quadrant, where both x and y are negative. Keeping this in mind can help you check your signs of the sine and cosine function. − 3 7π = −6 3 x = 12 cos = 12 2 6 7π −1 y = 12 sin = 12 = −6 6 2
The coordinates of the point are (−6 3 ,−6) . Try it Now 3. Find the coordinates of the point on a circle of radius 5 at an angle of
5π . 3
Example 7 We now have the tools to return to the sailboat question posed at the beginning of this section. A distress signal is sent from a sailboat during a storm, but the transmission is unclear and the rescue boat sitting at the marina cannot determine the sailboat’s location. Using high powered radar, they determine the distress signal is coming from a distance of 20 miles at an angle of 225 degrees from the marina. How many miles east/west and north/south of the rescue boat is the stranded sailboat? We can now answer the question by finding the coordinates of the point on a circle with a radius of 20 miles at an angle of 225 degrees. − 2 ≈ −14.142 miles x = 20 cos(225°) = 20 2 − 2 y = 20 sin (225°) = 20 ≈ −14.142 miles 2
N
225°
W 20 mi S
The sailboat is located 14.142 miles west and 14.142 miles south of the marina.
E
330 Chapter 5 The special values of sine and cosine in the first quadrant are very useful to know, since knowing them allows you to quickly evaluate the sine and cosine of very common angles without needing to look at a reference or use your calculator. However, scenarios do come up where we need to know the sine and cosine of other angles. To find the cosine and sine of any other angle, we turn to a computer or calculator. Be aware: most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When you evaluate “cos(30)” on your calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode. Most computer software with cosine and sine functions only operates in radian mode. Example 8 Evaluate the cosine of 20 degrees using a calculator or computer. On a calculator that can be put in degree mode, you can evaluate this directly to be approximately 0.939693. On a computer or calculator without degree mode, you would first need to convert the π angle to radians, or equivalently evaluate the expression cos 20 ⋅ . 180 Important Topics of This Section The sine function The cosine function Pythagorean Identity Unit Circle values Reference angles Using technology to find points on a circle Try it Now Answers 1. cos(π ) = −1 sin(π ) = 0 2.
π x = 3 cos = 3 ⋅ 0 = 0 2 π y = 3 sin = 3 ⋅ 1 = 3 2
5 −5 3 3. , 2 2
Section 5.3 Points on Circles Using Sine and Cosine 331
Section 5.3 Exercises 1. Find the quadrant in which the terminal point determined by t lies if a. sin(t ) < 0 and cos(t ) < 0 b. sin(t ) > 0 and cos(t ) < 0 2. Find the quadrant in which the terminal point determined by t lies if a. sin(t ) < 0 and cos(t ) > 0 b. sin(t ) > 0 and cos(t ) > 0 3. The point P is on the unit circle. If the y-coordinate of P is find the x coordinate. 4. The point P is on the unit circle. If the x-coordinate of P is IV, find the y coordinate.
3 , and P is in quadrant II, 5 1 , and P is in quadrant 5
1 and θ is in the 4th quadrant, find sin (θ ) . 7 2 6. If cos (θ ) = and θ is in the 1st quadrant, find sin (θ ) . 9 3 7. If sin (θ ) = and θ is in the 2nd quadrant, find cos (θ ) . 8 1 8. If sin (θ ) = − and θ is in the 3rd quadrant, find cos (θ ) . 4 5. If cos (θ ) =
9. For each of the following angles, find the reference angle and which quadrant the angle lies in. Then compute sine and cosine of the angle. a. 225° b. 300° c. 135° d. 210° 10. For each of the following angles, find the reference angle and which quadrant the angle lies in. Then compute sine and cosine of the angle. a. 120° b. 315° c. 250° d. 150° 11. For each of the following angles, find the reference angle and which quadrant the angle lies in. Then compute sine and cosine of the angle. 5π 5π 3π 7π a. b. c. d. 4 6 3 4 12. For each of the following angles, find the reference angle and which quadrant the angle lies in. Then compute sine and cosine of the angle. 4π 2π 5π 7π a. b. c. d. 3 3 6 4
332 Chapter 5 13. Give exact values for sin (θ ) and cos (θ ) for each of these angles. a. −
3π 4
b.
23π 6
c. −
π
d. 5π
2
14. Give exact values for sin (θ ) and cos (θ ) for each of these angles. a. −
2π 3
b.
17π 4
c. −
π 6
d. 10π
15. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same sine value as: 4π π b. 80° c. 140° d. e. 305° a. 3 3 16. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same sine value as: 7π π b. 15° c. 160° d. e. 340° a. 4 6 17. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same cosine value as: π 4π a. b. 80° c. 140° d. e. 305° 3 3 18. Find an angle θ with 0 < θ < 360° or 0 < θ < 2π that has the same cosine value as: π 7π a. b. 15° c. 160° d. e. 340° 6 4 19. Find the coordinates of the point on a circle with radius 15 corresponding to an angle of 220°. 20. Find the coordinates of the point on a circle with radius 20 corresponding to an angle of 280°. 21. Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12 seconds to reach the northernmost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive y-axis. [UW] a) Give Marla’s coordinates at her starting point. b) Give Marla’s coordinates when she has been running for 10 seconds. c) Give Marla’s coordinates when she has been running for 901.3 seconds.
Section 5.4 The Other Trigonometric Functions 333
Section 5.4 The Other Trigonometric Functions In the previous section, we defined the sine and cosine functions as ratios of the sides of a right triangle in a circle. Since the triangle has 3 sides there are 6 possible combinations of ratios. While the sine and cosine are the two prominent ratios that can be formed, there are four others, and together they define the 6 trigonometric functions. Tangent, Secant, Cosecant, and Cotangent Functions For the point (x, y) on a circle of radius r at an angle of θ , we can define four additional important functions as the ratios of the sides of the corresponding triangle: y The tangent function: tan(θ ) = x r The secant function: sec(θ ) = x r The cosecant function: csc(θ ) = y x The cotangent function: cot(θ ) = y
(x, y) r θ x
y
Geometrically, notice that the definition of tangent corresponds with the slope of the line segment between the origin (0, 0) and the point (x, y). This relationship can be very helpful in thinking about tangent values. You may also notice that the ratios defining the secant, cosecant, and cotangent are the reciprocals of the ratios defining the cosine, sine, and tangent functions, respectively. Additionally, notice that using our results from the last section, y r sin(θ ) sin(θ ) = tan(θ ) = = x r cos(θ ) cos(θ ) Applying this concept to the other trig functions we can state the other reciprocal identities. Identities The other four trigonometric functions can be related back to the sine and cosine functions using these basic relationships: tan(θ ) =
sin(θ ) cos(θ )
sec(θ ) =
1 cos(θ )
csc(θ ) =
1 sin(θ )
cot( = θ)
1 cos(θ ) = tan(θ ) sin(θ )
334 Chapter 5 These relationships are called identities. Identities are statements that are true for all values of the input on which they are defined. Identities are usually something that can be derived from definitions and relationships we already know, similar to how the identities above were derived from the circle relationships of the six trig functions. The Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine. We will discuss the role of identities more after an example. Example 1
5π Evaluate tan(45°) and sec . 6
Since we know the sine and cosine values for these angles, it makes sense to relate the tangent and secant values back to the sine and cosine values. sin( 45°) tan(45°) = = cos(45°)
2 2
2 =1 2
Notice this result is consistent with our interpretation of the tangent value as the slope of the line passing through the origin at the given angle: a line at 45 degrees would indeed have a slope of 1. 1 1 −2 −2 3 5π , which could also be written as . sec = = = 3 3 6 cos 5π − 3 2 6 Try it Now
7π 1. Evaluate csc . 6
Just as we often need to simplify algebraic expressions, it is often also necessary or helpful to simplify trigonometric expressions. To do so, we utilize the definitions and identities we have established.
Section 5.4 The Other Trigonometric Functions 335 Example 2 Simplify
sec(θ ) . tan (θ )
We can simplify this by rewriting both functions in terms of sine and cosine 1 sec(θ ) cos(θ ) To divide the fractions we could invert and multiply = tan (θ ) sin (θ ) cos(θ ) 1 cos(θ ) cancelling the cosines, = cos(θ ) sin (θ ) 1 simplifying and using the identity = = csc(θ ) sin (θ ) sec(θ ) can be simplified to csc(θ ) , we have, in fact, established a new tan (θ ) sec(θ ) identity: that = csc(θ ) . tan (θ ) By showing that
Occasionally a question may ask you to “prove the identity” or “establish the identity.” This is the same idea as when an algebra book asks a question like “show that ( x − 1) 2 = x 2 − 2 x + 1 .” In this type of question we must show the algebraic manipulations that demonstrate that the left and right side of the equation are in fact equal. You can think of a “prove the identity” problem as a simplification problem where you know the answer: you know what the end goal of the simplification should be, and just need to show the steps to get there. To prove an identity, in most cases you will start with the expression on one side of the identity and manipulate it using algebra and trigonometric identities until you have simplified it to the expression on the other side of the equation. Do not treat the identity like an equation to solve – it isn’t! The proof is establishing if the two expressions are equal, so we must take care to work with one side at a time rather than applying an operation simultaneously to both sides of the equation. Example 3 Prove the identity
1 + cot(α ) = sin(α ) + cos(α ) . csc(α )
Since the left side seems a bit more complicated, we will start there and simplify the expression until we obtain the right side. We can use the right side as a guide for what might be good steps to make. In this case, the left side involves a fraction while the right side doesn’t, which suggests we should look to see if the fraction can be reduced.
336 Chapter 5 Additionally, since the right side involves sine and cosine and the left does not, it suggests that rewriting the cotangent and cosecant using sine and cosine might be a good idea. 1 + cot(α ) csc(α ) cos(α ) 1+ sin(α ) = 1 sin(α )
Rewriting the cotangent and cosecant
To divide the fractions, we invert and multiply
cos(α ) sin(α ) = 1 + sin(α ) 1 sin(α ) cos(α ) sin(α ) 1⋅ = + ⋅ 1 sin(α ) 1 = sin(α ) + cos(α )
Distributing, Simplifying the fractions, Establishing the identity.
cos(α ) before sin(α ) inverting and multiplying. It is very common when proving or simplifying identities for there to be more than one way to obtain the same result. Notice that in the second step, we could have combined the 1 and
We can also utilize identities we have previously learned, like the Pythagorean Identity, while simplifying or proving identities. Example 4 Establish the identity
cos 2 (θ ) = 1 − sin (θ ) . 1 + sin (θ )
Since the left side of the identity is more complicated, it makes sense to start there. To simplify this, we will have to reduce the fraction, which would require the numerator to have a factor in common with the denominator. Additionally, we notice that the right side only involves sine. Both of these suggest that we need to convert the cosine into something involving sine. Recall the Pythagorean Identity told us cos 2 (θ ) + sin 2 (θ ) = 1 . By moving one of the trig functions to the other side, we can establish: sin 2 (θ ) = 1 − cos 2 (θ )
and
cos 2 (θ ) = 1 − sin 2 (θ )
Utilizing this, we now can establish the identity. We start on one side and manipulate:
Section 5.4 The Other Trigonometric Functions 337
cos 2 (θ ) 1 + sin (θ ) 1 − sin 2 (θ ) = 1 + sin (θ ) (1 − sin (θ ))(1 + sin (θ )) = 1 + sin (θ ) = 1 − sin (θ )
Utilizing the Pythagorean Identity Factoring the numerator Cancelling the like factors Establishing the identity
We can also build new identities from previously established identities. For example, if we divide both sides of the Pythagorean Identity by cosine squared (which is allowed since we’ve already shown the identity is true), cos 2 (θ ) + sin 2 (θ ) 1 Splitting the fraction on the left, = 2 cos (θ ) cos 2 (θ ) cos 2 (θ ) sin 2 (θ ) 1 + = 2 2 cos (θ ) cos (θ ) cos 2 (θ )
Simplifying and using the definitions of tan and sec
1 + tan 2 (θ ) = sec 2 (θ ) . Try it Now 2. Use a similar approach to establish that cot 2 (θ ) + 1 = csc 2 (θ ) . Identities Alternate forms of the Pythagorean Identity 1 + tan 2 (θ ) = sec 2 (θ ) cot 2 (θ ) + 1 = csc 2 (θ ) Example 5 If tan(θ ) =
2 and θ is in the 3rd quadrant, find cos(θ ) . 7
There are two approaches to this problem, both of which work equally well. Approach 1
y and the angle is in the third quadrant, we can imagine a triangle in a x circle of some radius so that the point on the circle is (-7, -2). Using the Pythagorean Theorem, we can find the radius of the circle: (−7) 2 + (−2) 2 = r 2 , so r = 53 .
Since tan(θ ) =
338 Chapter 5 Now we can find the cosine value: x −7 cos(θ ) = = r 53 Approach 2 Using the 1 + tan 2 (θ ) = sec 2 (θ ) form of the Pythagorean Identity with the known tangent value, 1 + tan 2 (θ ) = sec 2 (θ ) 2
2 1 + = sec 2 (θ ) 7 53 = sec 2 (θ ) 49 53 53 sec(θ ) = ± =± 49 7
Since the angle is in the third quadrant, the cosine value will be negative so the secant value will also be negative. Keeping the negative result, and using definition of secant, 53 sec(θ ) = − 7 1 53 Inverting both sides =− cos(θ ) 7 cos(θ ) = −
7 53
Try it Now 3. If sec(φ ) = −
=−
7 53 53
π 7 and < φ < π , find tan(φ ) and sin(φ ) . 3 2
Important Topics of This Section 6 Trigonometric Functions: Sine Cosine Tangent Cosecant Secant Cotangent Trig identities
Section 5.4 The Other Trigonometric Functions 339 Try it Now Answers 1. -2 2.
cos 2 (θ ) + sin 2 (θ ) =1 sin 2 θ cos 2 (θ ) sin 2 (θ ) 1 + = 2 2 sin (θ ) sin (θ ) sin 2 (θ ) cot 2 (θ ) + 1 = csc 2 (θ )
3. sin(φ ) =
40 7
tan(φ ) =
40 −3
340 Chapter 5
Section 5.4 Exercises 1. If θ =
π 4
, find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) .
2. If θ =
7π , find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 4
3. If θ =
5π , find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 6
4. If θ =
π 6
, find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) .
5. If θ =
2π , find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 3
6. If θ =
4π , find exact values for sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 3
7. Evaluate: a. sec (135° ) b. csc ( 210° ) c. tan ( 60° ) 8. Evaluate: a. sec ( 30° )
d. cot ( 225° )
b. csc ( 315° ) c. tan (135° ) d. cot (150° )
9. If sin (θ ) =
3 , and θ is in quadrant II, find cos (θ ) , sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 4
10. If sin (θ ) =
2 , and θ is in quadrant II, find cos (θ ) , sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 7
1 11. If cos (θ ) = − , and θ is in quadrant III, find 3 sin (θ ) , sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 1 12. If cos (θ ) = , and θ is in quadrant I, find sin (θ ) , sec (θ ) , csc (θ ) , tan (θ ) , cot (θ ) . 5
13. If tan (θ ) =
12 π , and 0 ≤ θ < , find sin (θ ) , cos (θ ) ,sec (θ ) , csc (θ ) , cot (θ ) . 2 5
14. If tan (θ ) = 4 , and 0 ≤ θ <
π 2
, find sin (θ ) , cos (θ ) ,sec (θ ) , csc (θ ) , cot (θ ) .
Section 5.4 The Other Trigonometric Functions 341 15. Use a calculator to find sine, cosine, and tangent of the following values: a. 0.15
b. 4
c. 70°
d. 283°
16. Use a calculator to find sine, cosine, and tangent of the following values: a. 0.5
b. 5.2
c. 10°
d. 195°
Simplify each of the following to an expression involving a single trig function with no fractions. 17. csc(t ) tan ( t ) 18. cos(t ) csc ( t ) 19. 20. 21. 22. 23. 24. 25. 26.
sec ( t )
csc ( t ) cot ( t )
csc ( t ) sec ( t ) − cos ( t ) sin ( t )
tan ( t )
sec ( t ) − cos ( t ) 1 + cot ( t )
1 + tan ( t ) 1 + sin ( t )
1 + csc ( t )
sin 2 ( t ) + cos 2 ( t ) cos 2 ( t )
1 − sin 2 ( t ) sin 2 ( t )
342 Chapter 5 Prove the identities. sin 2 (θ ) 27. = 1 − cos (θ ) 1 + cos (θ ) 1 −1 cos 2 ( t )
tan 2 (t ) 28. =
29. sec ( a ) − cos ( a ) = sin ( a ) tan ( a ) 30. 31. 32. 33.
1 + tan 2 ( b ) tan 2 ( b )
= csc 2 (b)
csc 2 ( x ) − sin 2 ( x ) csc ( x ) + sin ( x )
sin (θ ) − cos (θ ) sec (θ ) − csc (θ )
csc 2 (α ) − 1
= cos ( x ) cot ( x )
= sin (θ ) cos (θ )
csc 2 (α ) − csc (α )
= 1 + sin (α )
34. 1 + cot = ( x ) cos ( x ) ( sec ( x ) + csc ( x ) ) 35.
1 + cos ( u ) sin ( u )
2 36. 2sec = (t )
=
sin ( u )
1 − cos ( u )
1 − sin ( t ) cos ( t ) 2
+
1 1 − sin ( t )
sin 4 ( γ ) − cos 4 ( γ ) = sin ( γ ) + cos ( γ ) 37. sin ( γ ) − cos ( γ ) 38.
(1 + cos ( A) ) (1 − cos ( A) ) = sin sin ( A )
( A)
Section 5.5 Right Triangle Trigonometry 343
Section 5.5 Right Triangle Trigonometry In section 5.3 we were introduced to the sine and cosine function as ratios of the sides of a triangle drawn inside a circle, and spent the rest of that section discussing the role of those functions in finding points on the circle. In this section, we return to the triangle, and explore the applications of the trigonometric functions to right triangles where circles may not be involved. Recall that we defined sine and cosine as y sin(θ ) = r x cos(θ ) = r
(x, y) r θ x
y
Separating the triangle from the circle, we can make equivalent but more general definitions of the sine, cosine, and tangent on a right triangle. On the right triangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle. Right Triangle Relationships Given a right triangle with an angle of θ opposite hypotenuse adjacent cos(θ ) = hypotenuse opposite tan(θ ) = adjacent
sin(θ ) =
hypotenuse opposite θ adjacent
A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.” Example 1 Given the triangle shown, find the value for cos(α ) . The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so adjacent 15 = cos(α ) = hypotenuse 17
17
8
α 15
344 Chapter 5 When working with general right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the sine and cosine of either of the two acute angles in the triangle. Adjacent to β Opposite α
Adjacent to α Opposite β
β
α
Hypotenuse
Example 2 Using the triangle shown, evaluate cos(α ) , sin(α ) , cos( β ) , and sin( β ) .
4 3
adjacent to α 3 = hypotenuse 5 opposite α 4 sin(α ) = = hypotenuse 5 adjacent to β 4 cos( β ) = = hypotenuse 5 opposite β 3 sin( β ) = = hypotenuse 5 cos(α ) =
β
α 5
Try it Now 1. A right triangle is drawn with angle α opposite a side with length 33, angle β opposite a side with length 56, and hypotenuse 65. Find the sine and cosine of α and β . You may have noticed that in the above example that cos(α ) = sin( β ) and cos( β ) = sin(α ) . This makes sense since the side opposite α is also adjacent to β. Since the three angles in a triangle need to add to π, or 180 degrees, then the other two angles must add to
π
2
, or 90 degrees, so β =
π then cos(α ) = sin − α . 2
π
2
− α , and α =
π
2
− β . Since cos(α ) = sin( β ) ,
Section 5.5 Right Triangle Trigonometry 345 Cofunction Identities The cofunction identities for sine and cosine π π sin(θ ) = cos − θ cos(θ ) = sin − θ 2 2 In the previous examples we evaluated the sine and cosine on triangles where we knew all three sides of the triangle. Right triangle trigonometry becomes powerful when we start looking at triangles in which we know an angle but don’t know all the sides. Example 3 Find the unknown sides of the triangle pictured here. Since sin(θ ) =
opposite , hypotenuse
30°
7 sin(30°) = b From this, we can solve for the side b.
b
a
b sin(30°) = 7 7 b= sin(30°)
7
To obtain a value, we can evaluate the sine and simplify 7 b= = 14 1 2 To find the value for side a, we could use the cosine, or simply apply the Pythagorean Theorem: a2 + 72 = b2 a 2 + 7 2 = 14 2 a = 147 Notice that if we know at least one of the non-right angles of a right triangle and one side, we can find the rest of the sides and angles. Try it Now 2. A right triangle has one angle of sides and angles of the triangle.
π 3
and a hypotenuse of 20. Find the unknown
346 Chapter 5 Example 4 To find the height of a tree, a person walks to a point 30 feet from the base of the tree, and measures the angle from the ground to the top of the tree to be 57 degrees. Find the height of the tree. We can introduce a variable, h, to represent the height of the tree. The two sides of the triangle that are most important to us are the side opposite the angle, the height of the tree we are looking for, and the adjacent side, the side we are told is 30 feet long.
57° 30 feet
The trigonometric function which relates the side opposite of the angle and the side adjacent to the angle is the tangent. opposite h = adjacent 30 h = 30 tan(57°) h = 30 tan(57°) ≈ 46.2 feet tan(57°) =
Solving for h, Using technology we can approximate a value
The tree is approximately 46 feet tall. Example 5 A person standing on the roof of a 100 foot building is looking towards a skyscraper a few blocks away, wondering how tall it is. She measures the angle of declination from the roof of the building to the base of the skyscraper to be 20 degrees and the angle of inclination to the top of the skyscraper to be 42 degrees. To approach this problem, it would be good to start with a picture. Although we are interested in the height, h, of the skyscraper, it can be helpful to also label other unknown quantities in the picture – in this case the horizontal distance x between the buildings and a, the height of the skyscraper above the person.
a 42° 20°
x
100 ft To start solving this problem, notice we have two right triangles. In the top triangle, we know one angle is 42 degrees, but we don’t know any of the sides of the triangle, so we don’t yet know enough to work with this triangle. 100 ft
h
Section 5.5 Right Triangle Trigonometry 347 In the lower right triangle, we know one angle is 20 degrees, and we know the vertical height measurement of 100 ft. Since we know these two pieces of information, we can solve for the unknown distance x. opposite 100 Solving for x tan(20°) = = adjacent x x tan(20°) = 100 100 x= tan(20°) Now that we have found the distance x, we know enough information to solve the top right triangle. a opposite a = = tan(42°) = 100 adjacent x tan(20°) a tan(20°) tan(42°) = 100 100 tan(42°) = a tan(20°) 100 tan(42°) =a tan(20°) Approximating a value, 100 tan(42°) a= ≈ 247.4 feet tan(20°) Adding the height of the first building, we determine that the skyscraper is about 347 feet tall. Important Topics of This Section SOH CAH TOA Cofunction identities Applications with right triangles Try it Now Answers 56 33 56 33 1. sin(α ) = cos(α ) = sin( β ) = cos( β ) = 65 65 65 65 Adj π π adjacent 1 2. cos = so, adjacent = 20 cos = 20 = 10 = 3 2 3 hypoteuse 20 π Opposite Opp sin = = 20 3 hypoteuse
3 π = 10 3 so, opposite = 20 sin = 20 2 3
Missing angle = 30 degrees or π
6
348 Chapter 5
Section 5.5 Exercises Note: pictures may not be drawn to scale. In each of the triangles below, find sin ( A ) , cos ( A ) , tan ( A ) ,sec ( A ) , csc ( A ) , cot ( A ) . 1.
2.
A 4
10 8
A
10
In each of the following triangles, solve for the unknown sides and angles. 3. 4. B
10
c
7
5.
A
30°
b
a c
7.
6.
B
c
7
62°
b
8.
a
b
B 10
60°
c
10 A
a
65°
a
35°
12
B b
10°
9. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 10. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building?
Section 5.5 Right Triangle Trigonometry 349 11. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building. 12. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building. 13. A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36° and that the angle of depression to the bottom of the tower is 23°. How tall is the tower? 14. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 43° and that the angle of depression to the bottom of the tower is 31°. How tall is the tower? 15. A 200 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument? 16. A 400 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18° and that the angle of depression to the bottom of the tower is 3°. How far is the person from the monument? 17. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40°. From the same location, the angle of elevation to the top of the antenna is measured to be 43°. Find the height of the antenna. 18. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be 36°. From the same location, the angle of elevation to the top of the lightning rod is measured to be 38°. Find the height of the lightning rod. 19. Find the length x.
20. Find the length x.
82
85
63°
39° x
50°
36° x
350 Chapter 5 21. Find the length x.
22. Find the length x.
115
119 56°
35° x
70°
26° x
23. A plane is flying 2000 feet above sea level toward a mountain. The pilot observes the top of the mountain to be 18o above the horizontal, then immediately flies the plane at an angle of 20o above horizontal. The airspeed of the plane is 100 mph. After 5 minutes, the plane is directly above the top of the mountain. How high is the plane above the top of the mountain (when it passes over)? What is the height of the mountain? [UW] 24. Three airplanes depart SeaTac Airport. A Northwest flight is heading in a direction 50° counterclockwise from east, an Alaska flight is heading 115° counterclockwise from east and a Delta flight is heading 20° clockwise from east. Find the location of the Northwest flight when it is 20 miles north of SeaTac. Find the location of the Alaska flight when it is 50 miles west of SeaTac. Find the location of the Delta flight when it is 30 miles east of SeaTac. [UW]
Section 5.5 Right Triangle Trigonometry 351 25. The crew of a helicopter needs to land temporarily in a forest and spot a flat piece of ground (a clearing in the forest) as a potential landing site, but are uncertain whether it is wide enough. They make two measurements from A (see picture) finding α = 25° and β = 54°. They rise vertically 100 feet to B and measure γ = 47°. Determine the width of the clearing to the nearest foot. [UW] 26. A Forest Service helicopter needs to determine the width of a deep canyon. While hovering, they measure the angle γ = 48° at position B (see picture), then descend 400 feet to position A and make two measurements: α = 13° (the measure of ∠ EAD), β = 53° (the measure of ∠ CAD). Determine the width of the canyon to the nearest foot. [UW]
352 Chapter 5
Chapter 6: Periodic Functions
In the previous chapter, the trigonometric functions were introduced as ratios of sides of a right triangle, and related to points on a circle. We noticed how the x and y values of the points did not change with repeated revolutions around the circle by finding coterminal angles. In this chapter, we will take a closer look at the important characteristics and applications of these types of functions, and begin solving equations involving them. Section 6.1 Sinusoidal Graphs .................................................................................... 353 Section 6.2 Graphs of the Other Trig Functions ......................................................... 369 Section 6.3 Inverse Trig Functions ............................................................................. 379 Section 6.4 Solving Trig Equations ............................................................................ 387 Section 6.5 Modeling with Trigonometric Equations ................................................. 397
Section 6.1 Sinusoidal Graphs
The London Eye 1 is a huge Ferris wheel with diameter 135 meters (443 feet) in London, England, which completes one rotation every 30 minutes. When we look at the behavior of this Ferris wheel it is clear that it completes 1 cycle, or 1 revolution, and then repeats this revolution over and over again. This is an example of a periodic function, because the Ferris wheel repeats its revolution or one cycle every 30 minutes, and so we say it has a period of 30 minutes. In this section, we will work to sketch a graph of a rider’s height above the ground over time and express this height as a function of time. Periodic Functions A periodic function is a function for which a specific horizontal shift, P, results in the original function: f ( x + P) = f ( x) for all values of x. When this occurs we call the smallest such horizontal shift with P > 0 the period of the function. You might immediately guess that there is a connection here to finding points on a circle, since the height above ground would correspond to the y value of a point on the circle. We can determine the y value by using the sine function. To get a better sense of this function’s behavior, we can create a table of values we know, and use them to sketch a graph of the sine and cosine functions.
1
London Eye photo by authors, 2010, CC-BY
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
354 Chapter 6 Listing some of the values for sine and cosine on a unit circle, θ 0 3π π π π π 2π 4 6 4 3 2 3 cos 1 0 1 1 2 3 2 − − 2 2 2 2 2 sin 0 1 1 2 3 3 2 2 2 2 2 2
π
5π 6
− 1 2
3 2
-1 0
Here you can see how for each angle, we use the y value of the point on the circle to determine the output value of the sine function. f(θ) = sin(θ)
ππ π π 6 4 3
θ
2
Plotting more points gives the full shape of the sine and cosine functions. f(θ) = sin(θ)
θ
Notice how the sine values are positive between 0 and π, which correspond to the values of sine in quadrants 1 and 2 on the unit circle, and the sine values are negative between π and 2π, corresponding to quadrants 3 and 4.
Section 6.1 Sinusoidal Graphs 355 g(θ) = cos(θ)
θ
Like the sine function we can track the value of the cosine function through the 4 quadrants of the unit circle as we place it on a graph. Both of these functions are defined for all real numbers, since we can evaluate the sine and cosine of any angle. By thinking of sine and cosine as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1, 1] . Domain and Range of Sine and Cosine The domain of sine and cosine is all real numbers, (−∞, ∞) . The range of sine and cosine is the interval [-1, 1]. Both these graphs are called sinusoidal graphs. In both graphs, the shape of the graph begins repeating after 2π. Indeed, since any coterminal angles will have the same sine and cosine values, we could conclude that sin(θ + 2π ) = sin(θ ) and cos(θ + 2π ) = cos(θ ) . In other words, if you were to shift either graph horizontally by 2π, the resulting shape would be identical to the original function. Sinusoidal functions are a specific type of periodic function. Period of Sine and Cosine The periods of the sine and cosine functions are both 2π. Looking at these functions on a domain centered at the vertical axis helps reveal symmetries.
356 Chapter 6 sine
cosine
The sine function is symmetric about the origin, the same symmetry the cubic function has, making it an odd function. The cosine function is clearly symmetric about the y axis, the same symmetry as the quadratic function, making it an even function. Negative Angle Identities The sine is an odd function, symmetric about the origin, so sin(−θ ) = − sin(θ ) . The cosine is an even function, symmetric about the y-axis, so cos(−θ ) = cos(θ ) . These identities can be used, among other purposes, for helping with simplification and proving identities. π You may recall the cofunction identity from last chapter; sin(θ ) = cos − θ . 2 Graphically, this tells us that the sine and cosine graphs are horizontal transformations of each other. We can prove this by using the cofunction identity and the negative angle identity for cosine. π π π π sin(θ ) = cos − θ = cos − θ + = cos − θ − = cosθ − 2 2 2 2 Now we can clearly see that if we horizontally shift the cosine function to the right by π/2 we get the sine function. Remember this shift is not representing the period of the function. It only shows that the cosine and sine function are transformations of each other. Example 1 Simplify sin( −θ ) tan(θ ) − sin(θ ) tan(θ )
sin( −θ ) . tan(θ ) Using the even/odd identity Rewriting the tangent
Section 6.1 Sinusoidal Graphs 357
− sin(θ ) sin(θ ) cos(θ ) cos(θ ) − sin(θ ) ⋅ sin(θ ) − cos(θ )
Inverting and multiplying Simplifying we get
Transforming Sine and Cosine Example 2 A point rotates around a circle of radius 3. Sketch a graph of the y coordinate of the point. Recall that for a point on a circle of radius r, the y coordinate of the point is y = r sin(θ ) , so in this case, we get the equation y (θ ) = 3 sin(θ ) . The constant 3 causes a vertical stretch of the y values of the function by a factor of 3. Notice that the period of the function does not change. Since the outputs of the graph will now oscillate between -3 and 3, we say that the amplitude of the sine wave is 3. Try it Now 1. What is the amplitude of the function f (θ ) = 7 cos(θ ) ? Sketch a graph of this function. Example 3 A circle with radius 3 feet is mounted with its center 4 feet off the ground. The point closest to the ground is labeled P. Sketch a graph of the height above ground of the point P as the circle is rotated, then find a function that gives the height in terms of the angle of rotation.
3 ft 4 ft P
358 Chapter 6 Sketching the height, we note that it will start 1 foot above the ground, then increase up to 7 feet above the ground, and continue to oscillate 3 feet above and below the center value of 4 feet. Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. We decide to use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection. Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of one, so this graph has been vertically stretched by 3, as in the last example. Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, h(θ ) = −3 cos(θ ) + 4
Midline The center value of a sinusoidal function, the value that the function oscillates above and below, is called the midline of the function, corresponding to a vertical shift. The function f (θ ) = cos(θ ) + k has midline at y = k. Try it Now 2. What is the midline of the function f (θ ) = 3 cos(θ ) − 4 ? Sketch a graph of the function. To answer the Ferris wheel problem at the beginning of the section, we need to be able to express our sine and cosine functions at inputs of time. To do so, we will utilize composition. Since the sine function takes an input of an angle, we will look for a function that takes time as an input and outputs an angle. If we can find a suitable θ (t ) function, then we can compose this with our f (θ ) = cos(θ ) function to obtain a sinusoidal function of time: f (t ) = cos(θ (t )) .
Section 6.1 Sinusoidal Graphs 359 Example 4 A point completes 1 revolution every 2 minutes around a circle of radius 5. Find the x coordinate of the point as a function of time, if it starts at (5, 0). Normally, we would express the x coordinate of a point on a unit circle using x = r cos(θ ) , here we write the function x(θ ) = 5 cos(θ ) . The rotation rate of 1 revolution every 2 minutes is an angular velocity. We can use this rate to find a formula for the angle as a function of time. The point begins at an angle of 0. Since the point rotates 1 revolution x(θ) = 2π radians every 2 minutes, it rotates π radians every minute. After t minutes, it will have rotated: θ (t ) = π t radians θ Composing this with the cosine function, we obtain a function of time. x(t ) = 5 cos(θ (t )) = 5 cos(π t )
Notice that this composition has the effect of a horizontal compression, changing the period of the function. To see how the period relates to the stretch or compression coefficient B in the equation f (t ) = sin (Bt ) , note that the period will be the time it takes to complete one full revolution of a circle. If a point takes P minutes to complete 1 revolution, then the 2π radians 2π . Then θ (t ) = angular velocity is t . Composing with a sine function, P P minutes 2π f (t ) = sin(θ (t )) = sin t P
From this, we can determine the relationship between the coefficient B and the period: 2π . Notice that the stretch or compression coefficient B is a ratio of the “normal B= P period of a sinusoidal function” to the “new period.” If we know the stretch or 2π compression coefficient B, we can solve for the “new period”: P = . B
360 Chapter 6 Example 5
π What is the period of the function f (t ) = sin t ? 6
Using the relationship above, the stretch/compression factor is B = will be P =
2π 2π 6 = = 2π ⋅ = 12 . π B π 6
π 6
, so the period
While it is common to compose sine or cosine with functions involving time, the composition can be done so that the input represents any reasonable quantity. Example 6 A bicycle wheel with radius 14 inches has the bottom-most point on the wheel marked in red. The wheel then begins rolling down the street. Write a formula for the height above ground of the red point after the bicycle has travelled x inches. The height of the point begins at the lowest value, 0, increases to the highest value of 28 inches, and continues to oscillate above and below a center height of 14 inches. In terms of the angle of rotation, θ: h(θ ) = −14 cos(θ ) + 14 In this case, x is representing a linear distance the wheel has travelled, corresponding to an arclength along the circle. Since arclength and angle can be related by s = rθ , in this case we can write x = 14θ , which allows us to express the angle in terms of x: x θ ( x) = 14
Starting 14in
Rotated by θ θ x
Composing this with our cosine-based function from above, x 1 h( x) = h(θ ( x)) = −14 cos + 14 = −14 cos x + 14 14 14 2π 2π = = 2π ⋅ 14 = 28π , the circumference 1 B 14 of the circle. This makes sense – the wheel completes one full revolution after the bicycle has travelled a distance equivalent to the circumference of the wheel. The period of this function would be P =
Section 6.1 Sinusoidal Graphs 361 Summarizing our transformations so far: Transformations of Sine and Cosine Given an equation in the form f (t ) = A sin (Bt ) + k or f (t ) = A cos(Bt ) + k A is the vertical stretch, and is the amplitude of the function. B is the horizontal stretch/compression, and is related to the period, P, by P = k is the vertical shift and determines the midline of the function.
2π . B
P
y=k
A A P
Example 7 Determine the midline, amplitude, and period of the function f (t ) = 3 sin (2t ) + 1 . The amplitude is 3
2π 2π = =π B 2 The midline is at y = 1 The period is P =
Amplitude, midline, and period, when combined with vertical flips, allow us to write equations for a variety of sinusoidal situations. Try it Now 3. If a sinusoidal function starts on the midline at point (0,3), has an amplitude of 2, and a period of 4, write a formula for the function.
362 Chapter 6 Example 8 Find a formula for the sinusoidal function graphed here. The graph oscillates from a low of -1 to a high of 3, putting the midline at y = 1, halfway between. The amplitude will be 2, the distance from the midline to the highest value (or lowest value) of the graph. The period of the graph is 8. We can measure this from the first peak at x = -2 to the second at x = 6. Since the period is 8, the stretch/compression factor we will use will be 2π 2π π = = B= 4 8 P At x = 0, the graph is at the midline value, which tells us the graph can most easily be represented as a sine function. Since the graph then decreases, this must be a vertical reflection of the sine function. Putting this all together, π f (t ) = −2 sin t + 1 4 With these transformations, we are ready to answer the Ferris wheel problem from the beginning of the section. Example 9 The London Eye is a huge Ferris wheel with diameter 135 meters (443 feet) in London, England, which completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes. With a diameter of 135 meters, the wheel has a radius of 67.5 meters. The height will oscillate with amplitude of 67.5 meters above and below the center. Passengers board 2 meters above ground level, so the center of the wheel must be located 67.5 + 2 = 69.5 meters above ground level. The midline of the oscillation will be at 69.5 meters. The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with period of 30 minutes. Lastly, since the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a flipped cosine curve.
Section 6.1 Sinusoidal Graphs 363 Putting these together: Amplitude: 67.5 Midline: 69.5 2π π Period: 30, so B = = 30 15 Shape: -cos An equation for the rider’s height would be π h(t ) = −67.5 cos t + 69.5 15 Try it Now 4. The Ferris wheel at the Puyallup Fair 2 has a diameter of about 70 feet and takes 3 minutes to complete a full rotation. Passengers board from a platform 10 feet above the ground. Write an equation for a rider’s height above ground over time.
While these transformations are sufficient to represent many situations, occasionally we encounter a sinusoidal function that does not have a vertical intercept at the lowest point, highest point, or midline. In these cases, we need to use horizontal shifts. Recall that when the inside of the function is factored, it reveals the horizontal shift. Horizontal Shifts of Sine and Cosine Given an equation in the form f (t ) = A sin (B(t − h) ) + k or f (t ) = A cos(B(t − h) ) + k h is the horizontal shift of the function Example 10
π π Sketch a graph of f (t ) = 3 sin t − . 4 4 To reveal the horizontal shift, we first need to factor inside the function: π f (t ) = 3 sin (t − 1) 4
2
Photo by photogirl7.1, http://www.flickr.com/photos/kitkaphotogirl/432886205/sizes/z/, CC-BY
364 Chapter 6 This graph will have the shape of a sine function, starting at the midline and increasing, 2π 2π 4 with an amplitude of 3. The period of the graph will be P = = = 2π ⋅ = 8 . π B π 4 Finally, the graph will be shifted to the right by 1.
In some physics and mathematics books, you will hear the horizontal shift referred to as phase shift. In other physics and mathematics books, they would say the phase shift of the equation above is
π
, the value in the unfactored form. Because of this ambiguity, we 4 will not use the term phase shift any further, and will only talk about the horizontal shift. Example 11 Find a formula for the function graphed here. With highest value at 1 and lowest value at -5, the midline will be halfway between at -2. The distance from the midline to the highest or lowest value gives an amplitude of 3. The period of the graph is 6, which can be measured from the peak at x = 1 to the next peak at x = 7, or from the distance between the lowest points. This gives B =
2π 2π π = = . P 6 3
For the shape and shift, we have more than one option. We could either write this as: A cosine shifted 1 to the right A negative cosine shifted 2 to the left A sine shifted ½ to the left A negative sine shifted 2.5 to the right
Section 6.1 Sinusoidal Graphs 365 While any of these would be fine, the cosine shifts are easier to work with than the sine shifts in this case, because they involve integer values. Writing these: π or y ( x) = 3 cos ( x − 1) − 2 3 π y ( x) = −3 cos ( x + 2) − 2 3 Again, these functions are equivalent, so both yield the same graph. Try it Now 5. Write a formula for the function graphed here.
Important Topics of This Section Periodic functions Sine and cosine function from the unit circle Domain and range of sine and cosine functions Sinusoidal functions Negative angle identity Simplifying expressions Transformations Amplitude Midline Period Horizontal shifts Try it Now Answers 1. 7 2. -4 π 3. f ( x) 2sin x + 3 = 2 2π 4. h(t ) = −35cos t + 45 3 π π 5. Two possibilities: f= f ( x) 4sin ( x − 1) + 4 ( x) 4 cos ( x − 3.5) + 4 or= 5 5
366 Chapter 6
Section 6.1 Exercises
1. Sketch a graph of f ( x ) = −3sin ( x ) . 2. Sketch a graph of f ( x ) = 4sin ( x ) . 3. Sketch a graph of f ( x ) = 2 cos ( x ) . 4. Sketch a graph of f ( x ) = −4 cos ( x ) . For the graphs below, determine the amplitude, midline, and period, then find a formula for the function. 5.
6.
7.
8.
9.
10.
Section 6.1 Sinusoidal Graphs 367 For each of the following equations, find the amplitude, period, horizontal shift, and midline. 11.= y 3sin(8( x + 4)) + 5 π 12. y 4sin ( x − 3) + 7 = 2
13.= y 2sin(3 x − 21) + 4 14.= y 5sin(5 x + 20) − 2 π 15.= y sin x + π − 3 6 7π 7π 16. = y 8sin x+ 2 6
+6
Find a formula for each of the functions graphed below.
17.
18.
368 Chapter 6
19.
20. 21. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 50 degrees at midnight and the high and low temperature during the day are 57 and 43 degrees, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t. 22. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68 degrees at midnight and the high and low temperature during the day are 80 and 56 degrees, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D, in terms of t. 23. A Ferris wheel is 25 meters in diameter and boarded from a platform that is 1 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. The function h(t ) gives your height in meters above the ground t minutes after the wheel begins to turn. a. Find the amplitude, midline, and period of h ( t ) . b. Find a formula for the height function h ( t ) . c. How high are you off the ground after 5 minutes? 24. A Ferris wheel is 35 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 8 minutes. The function h(t ) gives your height in meters above the ground t minutes after the wheel begins to turn. a. Find the amplitude, midline, and period of h ( t ) . b. Find a formula for the height function h ( t ) . c. How high are you off the ground after 4 minutes?
Section 6.2 Graphs of the Other Trig Functions 369
Section 6.2 Graphs of the Other Trig Functions In this section, we will explore the graphs of the other four trigonometric functions. We’ll begin with the tangent function. Recall that in Chapter 5 we defined tangent as y/x or sine/cosine, so you can think of the tangent as the slope of a line through the origin making the given angle with the positive x axis. At an angle of 0, the line would be horizontal with a slope of zero. As the angle increases towards π/2, the slope increases more and more. At an angle of π/2, the line would be vertical and the slope would be undefined. Immediately past π/2, the line would have a steep negative slope, giving a large negative tangent value. There is a break in the function at π/2, where the tangent value jumps from large positive to large negative. We can use these ideas along with the definition of tangent to sketch a graph. Since tangent is defined as sine/cosine, we can determine that tangent will be zero when sine is zero: at -π, 0, π, and so on. Likewise, tangent will be undefined when cosine is zero: at -π/2, π/2, and so on. The tangent is positive from 0 to π/2 and π to 3π/2, corresponding to quadrants 1 and 3 of the unit circle. Using technology, we can obtain a graph of tangent on a standard grid. Notice that the graph appears to repeat itself. For any angle on the circle, there is a second angle with the same slope and tangent value halfway around the circle, so the graph repeats itself with a period of π; we can see one continuous cycle from - π/2 to π/2, before it jumps and repeats itself. The graph has vertical asymptotes and the tangent is undefined wherever a line at that angle would be vertical: at π/2, 3π/2, and so on. While the domain of the function is limited in this way, the range of the function is all real numbers. Features of the Graph of Tangent The graph of the tangent function m(θ ) = tan(θ ) The period of the tangent function is π The domain of the tangent function is θ ≠
π
+ kπ , where k is an integer 2 The range of the tangent function is all real numbers, (−∞, ∞)
370 Chapter 6 With the tangent function, like the sine and cosine functions, horizontal stretches/compressions are distinct from vertical stretches/compressions. The horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph. Example 1 Find a formula for the function graphed here. The graph has the shape of a tangent function, however the period appears to be 8. We can see one full continuous cycle from -4 to 4, suggesting a horizontal stretch. To stretch π to 8, the input values would have to be 8 multiplied by . Since the constant k
π in f (θ ) = a tan ( kθ ) is the reciprocal of the horizontal stretch
8
π
, the equation must have form
π f (θ ) = a tan θ . 8 We can also think of this the same way we did with sine and cosine. The period of the tangent function is π but it has been transformed and now it is 8; remember the ratio of
the “normal period” to the “new period” is
π
and so this becomes the value on the 8 inside of the function that tells us how it was horizontally stretched. To find the vertical stretch a, we can use a point on the graph. Using the point (2, 2) π π π 2 = a tan ⋅ 2 = a tan . Since tan = 1 , a = 2 4 4 8 π This function would have a formula f (θ ) = 2 tan θ . 8 Try it Now
π 1. Sketch a graph of f (θ ) = 3 tan θ . 6
Section 6.2 Graphs of the Other Trig Functions 371 1 . cos(θ ) Notice that the function is undefined when the cosine is 0, leading to a vertical asymptote in the graph at π/2, 3π/2, etc. Since the cosine is always no more than one in absolute value, the secant, being the reciprocal, will always be no less than one in absolute value. Using technology, we can generate the graph. The graph of the cosine is shown dashed so you can see the relationship. 1 f (θ ) = sec(θ ) = cos(θ ) For the graph of secant, we remember the reciprocal identity where sec(θ ) =
π The graph of cosecant is similar. In fact, since sin(θ ) = cos − θ , it follows that 2 π csc(θ ) = sec − θ , suggesting the cosecant graph is a horizontal shift of the secant 2 graph. This graph will be undefined where sine is 0. Recall from the unit circle that this occurs at 0, π, 2π, etc. The graph of sine is shown dashed along with the graph of the cosecant. 1 f (θ ) = csc(θ ) = sin(θ )
372 Chapter 6 Features of the Graph of Secant and Cosecant The secant and cosecant graphs have period 2π like the sine and cosine functions. Secant has domain θ ≠
π
+ kπ , where k is an integer 2 Cosecant has domain θ ≠ kπ , where k is an integer Both secant and cosecant have range of (−∞,−1] ∪ [1, ∞)
Example 2
π Sketch a graph of f (θ ) = 2 csc θ + 1 . What is the domain and range of this 2 function?
The basic cosecant graph has vertical asymptotes at the integer multiples of π. Because π 2 of the factor inside the cosecant, the graph will be compressed by , so the vertical 2 π 2 asymptotes will be compressed to θ = ⋅ kπ = 2k . In other words, the graph will have
π
vertical asymptotes at the integer multiples of 2, and the domain will correspondingly be θ ≠ 2k , where k is an integer. The basic sine graph has a range of [-1, 1]. The vertical stretch by 2 will stretch this to [-2, 2], and the vertical shift up 1 will shift the range of this function to [-1, 3]. The basic cosecant graph has a range of (−∞,−1] ∪ [1, ∞) . The vertical stretch by 2 will stretch this to (−∞,−2] ∪ [2, ∞) , and the vertical shift up 1 will shift the range of this function to (−∞,−1] ∪ [3, ∞) . Sketching a graph,
Notice how the graph of the transformed cosecant relates to the graph of π f (θ ) = 2 sin θ + 1 shown dashed. 2
Section 6.2 Graphs of the Other Trig Functions 373 Try it Now 2. Given the graph of π f (θ ) = 2 cos θ + 1 shown, sketch 2 π the graph of g (θ ) = 2 sec θ + 1 on 2 the same axes.
Finally, we’ll look at the graph of cotangent. Based on its definition as the ratio of cosine to sine, it will be undefined when the sine is zero: at at 0, π, 2π, etc. The resulting graph is similar to that of the tangent. In fact, it is a horizontal flip and shift of the tangent function, as we’ll see shortly in Example 3. 1 cos(θ ) f (θ ) = cot(θ ) = = tan(θ ) sin(θ )
Features of the Graph of Cotangent The cotangent graph has period π Cotangent has domain θ ≠ kπ , where k is an integer Cotangent has range of all real numbers, (−∞, ∞) In Section 6.1 we determined that the sine function was an odd function and the cosine was an even function by observing the graph and establishing the negative angle identities for cosine and sine. Similarly, you may notice from its graph that the tangent function appears to be odd. We can verify this using the negative angle identities for sine and cosine: sin (− θ ) − sin (θ ) tan (− θ ) = = = − tan (θ ) cos(− θ ) cos(θ ) The secant, like the cosine it is based on, is an even function, while the cosecant, like the sine, is an odd function.
374 Chapter 6 Negative Angle Identities Tangent, Cotangent, Secant and Cosecant tan (− θ ) = − tan (θ ) cot (− θ ) = − cot (θ )
sec(− θ ) = sec(θ )
csc(− θ ) = − csc(θ )
Example 3
π Prove that tan (θ ) = − cotθ − 2 tan (θ ) sin (θ ) = cos(θ ) π cos − θ 2 = π sin − θ 2 π = cot − θ 2 π = cot − θ − 2
Using the definition of tangent Using the cofunction identities
Using the definition of cotangent
Factoring a negative from the inside Using the negative angle identity for cot
π = − cotθ − 2 Important Topics of This Section The tangent and cotangent functions Period Domain Range The secant and cosecant functions Period Domain Range Transformations Negative Angle identities
Section 6.2 Graphs of the Other Trig Functions 375 Try it Now Answers
1.
2.
376 Chapter 6
Section 6.2 Exercises Match each trigonometric function with one of the graphs. 1. f ( x ) = tan ( x ) 2. f (x ) = sec( x ) 3. f ( x ) = csc( x)
4. f ( x ) = cot ( x )
I
II
III
IV
Find the period and horizontal shift of each of the following functions. 5. = f ( x ) 2 tan ( 4 x − 32 ) 6. = g ( x ) 3 tan ( 6 x + 42 ) π 7. h ( x ) 2sec ( x + 1) = 4 π 8. k ( x ) 3sec 2 x + = 2 π 9. m ( x ) 6 csc x + π = 3 20π 5π 10. n ( x ) 4 csc = x− 3 3
Section 6.2 Graphs of the Other Trig Functions 377 11. Sketch a graph of #7 above. 12. Sketch a graph of #8 above. 13. Sketch a graph of #9 above. 14. Sketch a graph of #10 above. π 15. Sketch a graph of j ( x ) = tan x . 2 π 16. Sketch a graph of= p ( t ) 2 tan t − . 2
Find a formula for each function graphed below.
17.
18.
19.
20.
378 Chapter 6 21. If tan x = −1.5 , find tan ( − x ) . 22. If tan x = 3 , find tan ( − x ) . 23. If sec x = 2 , find sec ( − x ) . 24. If sec x = −4 , find sec ( − x ) . 25. If csc x = −5 , find csc ( − x ) . 26. If csc x = 2 , find csc ( − x ) . Simplify each of the following expressions completely. 27. cot ( − x ) cos ( − x ) + sin ( − x ) 28. cos ( − x ) + tan ( − x ) sin ( − x )
Section 6.3 Inverse Trig Functions 379
Section 6.3 Inverse Trig Functions In previous sections we have evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that for a one-to-one function, if f (a ) = b , then an inverse function would satisfy f −1 (b) = a . You probably are already recognizing an issue – that the sine, cosine, and tangent functions are not one-to-one functions. To define an inverse of these functions, we will need to restrict the domain of these functions to yield a new function that is one-to-one. We choose a domain for each function that includes the angle zero.
π π Sine, limited to − , 2 2
Cosine, limited to [0, π ]
π π Tangent, limited to − , 2 2
On these restricted domains, we can define the inverse sine, inverse cosine, and inverse tangent functions. Inverse Sine, Cosine, and Tangent Functions π π For angles in the interval − , , if sin (θ ) = a , then sin −1 (a ) = θ 2 2 For angles in the interval [0, π ] , if cos(θ ) = a , then cos −1 (a ) = θ
π π For angles in the interval − , , if tan (θ ) = a , then tan −1 (a ) = θ 2 2 π π sin −1 ( x ) has domain [-1, 1] and range − , 2 2 −1 cos ( x ) has domain [-1, 1] and range [0, π ]
π π tan −1 ( x ) has domain of all real numbers and range − , 2 2
380 Chapter 6 The sin −1 ( x ) is sometimes called the arcsine function, and notated arcsin(a ) . The cos −1 ( x ) is sometimes called the arccosine function, and notated arccos(a ) . The tan −1 ( x ) is sometimes called the arctangent function, and notated arctan(a ) . The graphs of the inverse functions are shown here:
sin −1 ( x )
cos −1 ( x )
tan −1 ( x )
Notice that the output of each of these inverse functions is an angle. Example 1 Evaluate
1 a) sin −1 2
2 b) sin −1 − 2
3 c) cos −1 − 2
d) tan −1 (1)
1 1 a) Evaluating sin −1 is the same as asking what angle would have a sine value of . 2 2 1 In other words, what angle θ would satisfy sin (θ ) = ? There are multiple angles that 2 π 5π would satisfy this relationship, such as and , but we know we need the angle in 6 6 π π 1 π the interval − , , so the answer will be sin −1 = . Remember that the 2 2 2 6 inverse is a function so for each input, we will get exactly one output.
2 , we know that 5π and 7π both have a sine value of b) Evaluating sin −1 − 4 4 2
−
2 , but neither is in the interval 2
coterminal with
π π − 2 , 2 . For that, we need the negative angle
π 2 7π =− . . sin −1 − 4 4 2
Section 6.3 Inverse Trig Functions 381
3 , we are looking for an angle in the interval [0, π ] with a c) Evaluating cos −1 − 2 3 3 5π = cosine value of − . The angle that satisfies this is cos −1 − 6 . 2 2
π π d) Evaluating tan −1 (1) , we are looking for an angle in the interval − , with a 2 2 tangent value of 1. The correct angle is tan −1 (1) =
π
4
.
Try It Now 1. Evaluate a) sin −1 (− 1)
b) tan −1 (− 1)
c) cos −1 (− 1)
1 d) cos −1 2
Example 2 Evaluate sin −1 (0.97 ) using your calculator. Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode. In radian mode, sin −1 (0.97) ≈ 1.3252
In degree mode, sin −1 ( 0.97 ) ≈ 75.93°
Try it Now 2. Evaluate cos −1 (− 0.4 ) using your calculator. In Section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides. Example 3 Solve the triangle for the angle θ. Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.
12 θ
9
382 Chapter 6 9 Using the definition of the inverse, 12 9 θ = cos −1 Evaluating 12 θ ≈ 0.7227 , or about 41.4096° cos(θ ) =
There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions Example 4 13π Evaluate sin −1 cos 6
.
a) Here, we can directly evaluate the inside of the composition. 3 13π cos = 2 6 Now, we can evaluate the inverse function as we did earlier. 3 π = sin −1 2 3 Try it Now 11π 3. Evaluate cos −1 sin − 4
.
Example 5 4 Find an exact value for sin cos −1 . 5
4 Beginning with the inside, we can say there is some angle so θ = cos −1 , which 5 4 means cos(θ ) = , and we are looking for sin (θ ) . We can use the Pythagorean identity 5 to do this.
Section 6.3 Inverse Trig Functions 383
sin 2 (θ ) + cos 2 (θ ) = 1
Using our known value for cosine
2
4 sin 2 (θ ) + = 1 5 16 sin 2 (θ ) = 1 − 25 9 3 sin (θ ) = ± =± 25 5
Solving for sine
Since we know that the inverse cosine always gives an angle on the interval [0, π ] , we 3 4 know that the sine of that angle must be positive, so sin cos −1 = = θ) sin( 5 5 Example 6 7 Find an exact value for sin tan −1 . 4 While we could use a similar technique as in the last example, we will demonstrate a different technique here. From the inside, we 7 know there is an angle so tan (θ ) = . We can envision this as the 4 opposite and adjacent sides on a right triangle. Using the Pythagorean Theorem, we can find the hypotenuse of this triangle: 4 2 + 7 2 = hypotenuse 2
7 θ 4
hypotenuse = 65 Now, we can evaluate the sine of the angle as opposite side divided by hypotenuse 7 sin (θ ) = 65 This gives us our desired composition 7 7 . = θ) sin tan −1 = sin( 65 4 Try it Now 7 4. Evaluate cos sin −1 . 9
384 Chapter 6 We can also find compositions involving algebraic expressions. Example 7 x Find a simplified expression for cos sin −1 , for − 3 ≤ x ≤ 3 . 3 x . Using the Pythagorean Theorem, 3 Using our known expression for sine
We know there is an angle θ so that sin (θ ) =
sin 2 (θ ) + cos 2 (θ ) = 1 2
x 2 + cos (θ ) = 1 3 x2 2 cos (θ ) = 1 − 9
cos(θ ) = ±
Solving for cosine
9 − x2 9 − x2 =± 9 3
π π Since we know that the inverse sine must give an angle on the interval − , , we 2 2 can deduce that the cosine of that angle must be positive. This gives us x cos sin −1 = 3
Try it Now
9 − x2 3
(
)
5. Find a simplified expression for sin tan −1 (4 x ) , for −
1 1 ≤x≤ . 4 4
Important Topics of This Section Inverse trig functions: arcsine, arccosine and arctangent Domain restrictions Evaluating inverses using unit circle values and the calculator Simplifying numerical expressions involving the inverse trig functions Simplifying algebraic expressions involving the inverse trig functions
Section 6.3 Inverse Trig Functions 385 Try it Now Answers 1. a) −
π
2
b) −
π 4
2. 1.9823 or 113.578° 3π 4 4 2 4. 9 4x 5. 16 x 2 + 1
3.
c) π
d)
π 3
386 Chapter 6
Section 6.3 Exercises Evaluate the following expressions. 2 3 1. sin −1 2. sin −1 2 2 2 1 5. cos −1 6. cos −1 2 2 9. tan −1 (1)
10. tan −1
( 3)
2 7. cos −1 − 2
4. sin −1 − 8. cos −1 −
11. tan −1 − 3
12. tan −1 ( −1)
1 3. sin −1 − 2
(
)
Use your calculator to evaluate each expression. 14. cos −1 (0.8) 15. sin −1 (− 0.8) 13. cos −1 (− 0.4 )
2 2 3 2
16. tan −1 (6 )
Find the angle θ. 10
7
θ 17.
θ 18.
Evaluate the following expressions. π 19. sin −1 cos 4 4π 21. sin −1 cos 3 3 23. cos sin −1 7
(
)
12 19
π 20. cos −1 sin 6 5π 22. cos −1 sin 4 4 24. sin cos −1 9 1 26. tan sin −1 3
25. cos tan −1 (4 )
Find a simplified expression for each of the following. x x 27. sin cos −1 , for − 5 ≤ x ≤ 5 28. tan cos −1 , for − 2 ≤ x ≤ 2 2 5
(
)
29. sin tan −1 (3 x )
(
)
30. cos tan −1 (4 x )
Section 6.4 Solving Trig Equations 387
Section 6.4 Solving Trig Equations In Section 6.1, we determined the height of a rider on the London Eye Ferris wheel could π be determined by the equation h(t ) = −67.5 cos t + 69.5 . 15 If we wanted to know length of time during which the rider is more than 100 meters above ground, we would need to solve equations involving trig functions. Solving using known values In the last chapter, we learned sine and cosine values at commonly encountered angles. We can use these to solve sine and cosine equations involving these common angles. Example 1 Solve sin (t ) =
1 for all possible values of t. 2
1 . While 2 evaluating a function always produces one result, solving for an input can yield multiple
Notice this is asking us to identify all angles, t, that have a sine value of t =
solutions. Two solutions should immediately jump to mind from the last chapter: t =
π
5π and t = because they are the common angles on the unit circle. 6
6
Looking at a graph confirms that there are more than these two solutions. While eight are seen on this graph, there are an infinite number of solutions!
Remember that any coterminal angle will also have the same sine value, so any angle coterminal with these two is also a solution. Coterminal angles can be found by adding full rotations of 2π, so we end up with a set of solutions: π 5π t = + 2πk where k is an integer, and t = + 2πk where k is an integer 6 6
388 Chapter 6 Example 2 A circle of radius 5 2 intersects the line x = -5 at two points. Find the angles θ on the interval 0 ≤ θ < 2π , where the circle and line intersect. The x coordinate of a point on a circle can be found as x = r cos(θ ) , so the x coordinate of points on this circle would be x = 5 2 cos(θ ) . To find where the line x = -5 intersects the circle, we can solve for where the x value on the circle would be -5 Isolating the cosine − 5 = 5 2 cos(θ ) −1 2
= cos(θ )
cos(θ ) =
Recall that
−1 2
=
− 2 , so we are solving 2
− 2 2
We can recognize this as one of our special cosine values from our unit circle, and it corresponds with angles 5π 3π and θ = θ= 4 4 Try it Now 1. Solve tan ( t ) = 1 for all possible values of t. Example 3 The depth of water at a dock rises and falls with the tide, following the equation π f (t ) = 4 sin t + 7 , where t is measured in hours after midnight. A boat requires a 12 depth of 9 feet to tie up at the dock. Between what times will the depth be 9 feet? To find when the depth is 9 feet, we need to solve f(t) = 9. π 4 sin t + 7 = 9 Isolating the sine 12 π 4 sin t = 2 Dividing by 4 12 1 π π 1 sin t = We know sin (θ ) = when θ = 2 6 12 2
or θ =
5π 6
1 , because of the horizontal 2 stretch/compression, it is less clear how to proceed.
While we know what angles have a sine value of
Section 6.4 Solving Trig Equations 389 To deal with this, we can make a substitution, defining a new temporary variable u to be u=
π
12
t , so our equation becomes
1 2 From earlier, we saw the solutions to this equation were sin (u ) =
π
+ 2πk where k is an integer, and 6 5π u= + 2πk where k is an integer 6 u=
Undoing our substitution, we can replace the u in the solutions with u = for t.
π 12
t=
π 6
+ 2πk where k is an integer, and
π 12
t=
π 12
t and solve
5π + 2πk where k is an integer. 6
Dividing by π/12, we obtain solutions
t = 2 + 24k where k is an integer, and t = 10 + 24k where k is an integer. The depth will be 9 feet and the boat will be able to approach the dock between 2am and 10am. Notice how in both scenarios, the 24k shows how every 24 hours the cycle will be repeated.
π 1 In the previous example, looking back at the original simplified equation sin t = , 12 2 we can use the ratio of the “normal period” to the stretch factor to find the period. 2π 12 = 2π = 24 ; notice that the sine function has a period of 24, which is reflected π π 12 in the solutions: there were two unique solutions on one full cycle of the sine function, and additional solutions were found by adding multiples of a full period. Try it Now 2. Solve 4 sin (5t ) − 1 = 1 for all possible values of t.
390 Chapter 6 Solving using the inverse trig functions The solutions to sin (θ ) = 0.3 do not involve any of the “special” values of the trig functions to we have learned. To find the solutions, we need to use the inverse sine function. Example 4 Use the inverse sine function to find one solution to sin (θ ) = 0.8 . Since this is not a known unit circle value, calculating the inverse, θ = sin −1 (0.8) . This requires a calculator and we must approximate a value for this angle. If your calculator is in degree mode, your calculator will give you an angle in degrees as the output. If your calculator is in radian mode, your calculator will give you an angle in radians. In θ sin −1 ( 0.8 ) ≈ 53.130° . radians, θ = sin −1 (0.8) ≈ 0.927 , or in degrees, = If you are working with a composed trig function and you are not solving for an angle, you will want to ensure that you are working in radians. In calculus, we will almost always want to work with radians since they are unit-less. Notice that the inverse trig functions do exactly what you would expect of any function – for each input they give exactly one output. While this is necessary for these to be a function, it means that to find all the solutions to an equation like sin (θ ) = 0.8 , we need to do more than just evaluate the inverse function. Example 5 Find all solutions to sin (θ ) = 0.8 . We would expect two unique angles on one cycle to have 0.8 this sine value. In the previous example, we found one θ solution to be θ = sin −1 (0.8) ≈ 0.927 . To find the other, we 0.929 need to answer the question “what other angle has the same 1 sine value as an angle of 0.927?” On a unit circle, we would recognize that the second angle would have the same reference angle and reside in the second quadrant. This second angle would be located at θ = π − sin −1 (0.8) , or approximately θ ≈ π − 0.927 = 2.214 To find more solutions we recall that angles coterminal with these two would have the same sine value, so we can add full cycles of 2π.
θ = sin −1 (0.8) + 2πk and θ = π − sin −1 (0.8) + 2πk where k is an integer, or approximately, θ = 0.927 + 2πk and θ = 2.214 + 2πk where k is an integer.
Section 6.4 Solving Trig Equations 391 Example 6 Find all solutions to sin ( x ) = −
8 on the interval 0° ≤ x < 360° . 9
First we will turn our calculator to degree mode. Using the inverse, we can find one 8 solution x = sin −1 − ≈ −62.734° . While this angle satisfies the equation, it does not 9 lie in the domain we are looking for. To find the angles in the desired domain, we start looking for additional solutions. First, an angle coterminal with − 62.734° will have the same sine. By adding a full rotation, we can find an angle in the desired domain with the same sine. x = −62.734° + 360° = 297.266° There is a second angle in the desired domain that lies in the third quadrant. Notice that 62.734° is the reference angle for all solutions, so this second solution would be 62.734° past 180° x = 62.734° + 180° = 242.734° The two solutions on 0° ≤ x < 360° are x = 297.266° and x = 242.734° Example 7 Find all solutions to tan ( x ) = 3 on 0 ≤ x < 2π . Using the inverse tangent function, we can find one solution x = tan −1 (3) ≈ 1.249 . Unlike the sine and cosine, the tangent function only attains any output value once per cycle, so there is no second solution in any one cycle. By adding π, a full period of tangent function, we can find a second angle with the same tangent value. If additional solutions were desired, we could continue to add multiples of π, so all solutions would take on the form x = 1.249 + kπ , however we are only interested in 0 ≤ x < 2π . x = 1.249 + π = 4.391 The two solutions on 0 ≤ x < 2π are x = 1.249 and x = 4.391. Try it Now 3. Find all solutions to tan ( x ) = 0.7 on 0° ≤ x < 360° .
1.249
4.391
1
392 Chapter 6 Example 8 Solve 3 cos(t ) + 4 = 2 for all solutions on one cycle, 0 ≤ t < 2π Isolating the cosine 3 cos(t ) + 4 = 2 3 cos(t ) = −2 2 Using the inverse, we can find one solution cos(t ) = − 3 2 t = cos −1 − ≈ 2.301 3 Thinking back to the circle, a second angle with the same cosine would be located in the third quadrant. Notice that the location of this angle could be represented as t = −2.301 . To represent this as a positive angle we could find a coterminal angle by adding a full cycle. t = −2.301 + 2π = 3.982
−
2 3
2.301
-2.301 or 3.982
1
The equation has two solutions between 0 and 2π, at t = 2.301 and t = 3.982. Example 9 Solve cos(3t ) = 0.2 for all solutions on two cycles, 0 ≤ t <
4π . 3
As before, with a horizontal compression it can be helpful to make a substitution, u = 3t . Making this substitution simplifies the equation to a form we have already solved. cos(u ) = 0.2
u = cos −1 (0.2 ) ≈ 1.369
A second solution on one cycle would be located in the fourth quadrant with the same reference angle. u = 2π − 1.369 = 4.914 In this case, we need all solutions on two cycles, so we need to find the solutions on the second cycle. We can do this by adding a full rotation to the previous two solutions. u = 1.369 + 2π = 7.653 u = 4.914 + 2π = 11.197 Undoing the substitution, we obtain our four solutions: 3t = 1.369, so t = 0.456 3t = 4.914 so t = 1.638 3t = 7.653, so t = 2.551 3t = 11.197, so t = 3.732
Section 6.4 Solving Trig Equations 393 Example 10 Solve 3 sin (π t ) = −2 for all solutions. Isolating the sine 3 sin (π t ) = −2 2 We make the substitution u = π t sin (π t ) = − 3 2 Using the inverse, we find one solution sin (u ) = − 3 2 u = sin −1 − ≈ −0.730 3 This angle is in the fourth quadrant. A second angle with the same sine would be in the third quadrant with 0.730 as a reference angle: u = π + 0.730 = 3.871 We can write all solutions to the equation sin (u ) = − u = −0.730 + 2π k where k is an integer, or u = 3.871 + 2π k
2 as 3
Undoing our substitution, we can replace u in our solutions with u = π t and solve for t Divide by π π t = −0.730 + 2π k or π t = 3.871 + 2π k or t = −0.232 + 2k t = 1.232 + 2k Try it Now
π 4. Solve 5 sin t + 3 = 0 for all solutions on one cycle, 0 ≤ t < 2π . 2
Solving Trig Equations 1) Isolate the trig function on one side of the equation 2) Make a substitution for the inside of the sine, cosine, or tangent (or other trig function) 3) Use inverse trig functions to find one solution 4) Use symmetries to find a second solution on one cycle (when a second exists) 5) Find additional solutions if needed by adding full periods 6) Undo the substitution
We now can return to the question we began the section with.
394 Chapter 6 Example 11 The height of a rider on the London Eye Ferris wheel can be determined by the equation π h(t ) = −67.5 cos t + 69.5 . How long is the rider more than 100 meters above 15 ground? To find how long the rider is above 100 meters, we first find the times at which the rider is at a height of 100 meters by solving h(t) = 100. π 100 = −67.5 cos t + 69.5 Isolating the cosine 15 π 30.5 = −67.5 cos t 15 30.5 π π = cos t We make the substitution u = t − 67.5 15 15 30.5 Using the inverse, we find one solution = cos(u ) − 67.5
30.5 u = cos −1 ≈ 2.040 − 67.5 This angle is in the second quadrant. A second angle with the same cosine would be symmetric in the third quadrant. This angle could be represented as u = -2.040, but we need a coterminal positive angle, so we add 2π: u = 2π − 2.040 ≈ 4.244
u = 2.040
u = -2.040 or 4.244
Now we can undo the substitution to solve for t
π
15
π
15
t = 2.040 so t = 9.740 minutes after the start of the ride
t = 4.244 so t = 20.264 minutes after the start of the ride
A rider will be at 100 meters after 9.740 minutes, and again after 20.264. From the behavior of the height graph, we know the rider will be above 100 meters between these times. A rider will be above 100 meters for 20.265-9.740 = 10.523 minutes of the ride. Important Topics of This Section Solving trig equations using known values Using substitution to solve equations Finding answers in one cycle or period vs. finding all possible solutions Method for solving trig equations
1
Section 6.4 Solving Trig Equations 395 Try it Now Answers 1.
π
4
+πk
π 2π 2π k t= + k 6 5 30 5 3. x = 34.992° or x = 180° + 34.99° = 214.992° 4. t = 3.590 or t = 2.410 2. t =
π
+
396 Chapter 6
Section 6.4 Exercises Find all solutions on the interval 0 ≤ θ < 2π . 1. 2sin (θ ) = − 2 2. 2sin (θ ) = 3 3. 2 cos (θ ) = 1
4. 2 cos (θ ) = − 2
5. sin (θ ) = 1
6. sin (θ ) = 0
7. cos (θ ) = 0
8. cos (θ ) = −1
Find all solutions. 9. 2 cos (θ ) = 2
10. 2 cos (θ ) = −1
11. 2sin (θ ) = −1
12. 2sin (θ ) = − 3
Find all solutions. 13. 2sin ( 3θ ) = 1
14. 2sin ( 2θ ) = 3
15. 2sin ( 3θ ) = − 2
16. 2sin ( 3θ ) = −1
17. 2 cos ( 2θ ) = 1
18. 2 cos ( 2θ ) = 3
19. 2 cos ( 3θ ) = − 2
20. 2 cos ( 2θ ) = −1
π 22. sin θ = −1 3
23. 2sin (πθ ) = 1 .
π 21. cos θ = −1 4 π 24. 2 cos θ = 3 5
Find all solutions on the interval 0 ≤ x < 2π . − 26. sin ( x ) = 0.48 27. sin ( x ) = 0.58 25. sin ( x ) = 0.27
28. sin ( x ) = −0.34
29. cos ( x ) = −0.55
31. cos ( x ) = 0.71
32. cos ( x ) = −0.07
Find the first two positive solutions. 33. 7 sin ( 6 x ) = 2 34. 7 sin ( 5 x ) = 6
35. 5cos ( 3 x ) = −3
36. 3cos ( 4 x ) = 2
π 37. 3sin 4
π 39. 5cos 3
π 40. 3cos 2
x = 2
30. sin ( x ) = 0.28
π 38. 7 sin 5
x = 6
x =1
x = −2
Section 6.5 Modeling with Trigonometric Functions 397
Section 6.5 Modeling with Trigonometric Functions Solving right triangles for angles In Section 5.5, we used trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides. Example 1 An airplane needs to fly to an airfield located 300 miles east and 200 miles north of its current location. At what heading should the airplane fly? In other words, if we ignore air resistance or wind speed, how many degrees north of east should the airplane fly? We might begin by drawing a picture and labeling all of the known information. Drawing a triangle, we see we are looking for the angle α. In this triangle, the side opposite the angle α is 200 miles and the side adjacent is 300 miles. Since we know the values for the opposite and adjacent sides, it makes sense to use the tangent function. 200 Using the inverse, tan(α ) = 300 200 α = tan −1 ≈ 0.588 , or equivalently about 33.7 degrees. 300
α
200 300
The airplane needs to fly at a heading of 33.7 degrees north of east. Example 2 OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length 3. Find the angle such a ladder forms with the ground. For any length of ladder, the base needs to be one quarter of the distance the foot of the ladder is away from the wall. Equivalently, if the base is a feet from the wall, the ladder can be 4a feet long. Since a is the side adjacent to the angle and 4a is the hypotenuse, we use the cosine function. a 1 Using the inverse cos(θ ) = = 4a 4 1 4 The ladder forms a 75.52 degree angle with the ground.
θ = cos −1 ≈ 75.52 degrees
3
http://www.osha.gov/SLTC/etools/construction/falls/4ladders.html
4a θ a
398 Chapter 6 Try it Now 1. One of the cables that anchor the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground and the second anchor on the ground is 23 meters from the base of the Ferris wheel. What is the angle of elevation (from ground up to the center of the Ferris wheel) and how long is the cable? Example 3 In a video game design, a map shows the location of other characters relative to the player, who is situated at the origin, and the direction they are facing. A character currently shows on the map at coordinates (-3, 5). If the player rotates counterclockwise by 20 degrees, then the objects in the map will correspondingly rotate 20 degrees clockwise. Find the new coordinates of the character. To rotate the position of the character, we can imagine it as a point on a circle, and we will change the angle of the point by 20 degrees. To do so, we first need to find the radius of this circle and the original angle. Drawing a right triangle inside the circle, we can find the radius using the Pythagorean Theorem: 2 r2 ( −3) + 52 = r = 9 + 25 = 34
To find the angle, we need to decide first if we are going to find the acute angle of the triangle, the reference angle, or if we are going to find the angle measured in standard position. While either approach will work, in this case we will do the latter. Since for any point on a circle we know x = r cos(θ ) , using our given information we get − 3 = 34 cos(θ ) −3 = cos(θ ) 34 −3 θ = cos −1 ≈ 120.964° 34 While there are two angles that have this cosine value, the angle of 120.964 degrees is in the second quadrant as desired, so it is the angle we were looking for.
Rotating the point clockwise by 20 degrees, the angle of the point will decrease to 100.964 degrees. We can then evaluate the coordinates of the rotated point x = 34 cos(100.964°) ≈ −1.109 y = 34 sin(100.964°) ≈ 5.725
The coordinates of the character on the rotated map will be (-1.109, 5.725).
Section 6.5 Modeling with Trigonometric Functions 399 Modeling with sinusoidal functions Many modeling situations involve functions that are periodic. Previously we learned that sinusoidal functions are a special type of periodic function. Problems that involve quantities that oscillate can often be modeled by a sine or cosine function and once we create a suitable model for the problem we can use that model to answer various questions. Example 4 The hours of daylight in Seattle oscillate from a low of 8.5 hours in January to a high of 16 hours in July4. When should you plant a garden if you want to do it during a month where there are 14 hours of daylight? To model this, we first note that the hours of daylight oscillate with a period of 12 months. With a low of 8.5 and a high of 16, the midline will be halfway between these 16 + 8.5 values, at = 12.25 . The amplitude will be half the difference between the 2 16 − 8.5 highest and lowest values: = 3.75 , or equivalently the distance from the 2 midline to the high or low value, 16-12.25=3.75. Letting January be t = 0, the graph starts at the lowest value, so it can be modeled as a flipped cosine graph. Putting this together, we get a model: π h(t ) = −3.75 cos t + 12.25 3.75 y=12.25 6 -cos(t) represents the flipped cosine, 3.75 is the amplitude, 12.25 is the midline, 2π π = corresponds to the horizontal stretch, 12 6 found by using the ratio of the “original period / new period” h(t) is our model for hours of day light t months after January. To find when there will be 14 hours of daylight, we solve h(t) = 14. π 14 = −3.75 cos t + 12.25 6 π 1.75 = −3.75 cos t 6 1.75 π − = cos t 3.75 6 4
Isolating the cosine Subtracting 12.25 and dividing by -3.75 Using the inverse
http://www.mountaineers.org/seattle/climbing/Reference/DaylightHrs.html
400 Chapter 6
π
1.75 t = cos −1 − ≈ 2.0563 6 3.75 6 t = 2.0563 ⋅ = 3.927
multiplying by the reciprocal t=3.927 months past January
π
There will be 14 hours of daylight 3.927 months into the year, or near the end of April. While there would be a second time in the year when there are 14 hours of daylight, since we are planting a garden, we would want to know the first solution, in spring, so we do not need to find the second solution in this case.
Try it Now 2. The author’s monthly gas usage (in therms) is shown here. Find a function to model the data.
160 140 120 100 80 60 40 20 0 Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Example 6 An object is connected to the wall with a spring that has a natural length of 20 cm. The object is pulled back 8 cm past the natural length and released. The object oscillates 3 times per second. Find an equation for the horizontal position of the object ignoring the effects of friction. How much time during each cycle is the object more than 27 cm from the wall? If we use the distance from the wall, x, as the desired output, then the object will oscillate equally on either side of the spring’s natural length of 20, putting the midline of the function at 20 cm. If we release the object 8 cm past the natural length, the amplitude of the oscillation will be 8 cm. We are beginning at the largest value and so this function can most easily be modeled using a cosine function. Since the object oscillates 3 times per second, it has a frequency of 3 and the period of one oscillation is 1/3 of second. Using this we find the horizontal compression using the 2π ratios of the periods: = 6π . 1/ 3
Section 6.5 Modeling with Trigonometric Functions 401 Using all this, we can build our model: x(t ) = 8 cos(6π t ) + 20 To find when the object is 27 cm from the wall, we can solve x(t) = 27 Isolating the cosine 27 = 8 cos(6π t ) + 20 7 = 8 cos(6π t ) 7 Using the inverse = cos(6π t ) 8 7 6π t = cos −1 ≈ 0.505 8 0.505 t= = 0.0268 6π Based on the shape of the graph, we can conclude that the object will spend the first 0.0268 seconds more than 27 cm from the wall. Based on the symmetry of the function, the object will spend another 0.0268 seconds more than 27 cm from the wall at the end of the cycle. Altogether, the object spends 0.0536 seconds each cycle at a distance greater than 27 cm from the wall.
In some problems, we can use trigonometric functions to model behaviors more complicated than the basic sinusoidal function. Example 7 A rigid rod with length 10 cm is attached to a circle of radius 4cm at point A as shown here. The point B is able to freely move along the horizontal axis, driving a piston 5. If the wheel rotates counterclockwise at 5 revolutions per second, find the location of point B as a function of time. When will the point B be 12 cm from the center of the circle?
4cm θ
A
10 cm
B
To find the position of point B, we can begin by finding the coordinates of point A. Since it is a point on a circle with radius 4, we can express its coordinates as (4 cos(θ ),4 sin(θ )) , where θ is the angle shown. 5
For an animation of this situation, see http://www.mathdemos.org/mathdemos/sinusoidapp/engine1.gif
402 Chapter 6 The angular velocity is 5 revolutions per second, or equivalently 10π radians per second. After t seconds, the wheel will rotate by θ = 10π t radians. Substituting this, we can find the coordinates of A in terms of t. (4 cos(10π t ),4 sin(10π t )) Notice that this is the same value we would have obtained by observing that the period of the rotation is 1/5 of a second and calculating the stretch/compression factor: " original" 2π = 10π . " new" 1 5 Now that we have the coordinates of the point A, we can relate this to the point B. By drawing a vertical line segment from A to the horizontal axis, we can form a right triangle. The height of the triangle is the y coordinate of the point A: 4sin(10π t ) . Using the Pythagorean Theorem, we can find the base length of the triangle:
( 4sin(10π t ) )
2
A
10 cm b
B
+ b2 = 102
= b 2 100 − 16sin 2 (10π t ) = b
100 − 16sin 2 (10π t )
Looking at the x coordinate of the point A, we can see that the triangle we drew is shifted to the right of the y axis by 4 cos(10π t ) . Combining this offset with the length of the base of the triangle gives the x coordinate of the point B: = x(t ) 4 cos(10π t ) + 100 − 16sin 2 (10π t )
To solve for when the point B will be 12 cm from the center of the circle, we need to solve x(t) = 12. = 12 4 cos(10π t ) + 100 − 16sin 2 (10π t )
Isolate the square root
12 − 4 cos(10π t ) =− 100 16sin 2 (10π t )
Square both sides
(12 − 4 cos(10π t ) )
Expand the left side
2
100 − 16sin 2 (10π t ) =
144 − 96 cos(10π t ) + 16 cos 2 (10π t ) = 100 − 16sin 2 (10π t ) Move all terms to the left Factor out 16 44 − 96 cos(10π t ) + 16 cos 2 (10π t ) + 16sin 2 (10π t ) = 0 44 − 96 cos(10π t ) + 16 ( cos 2 (10π t ) + sin 2 (10π t ) ) = 0
At this point, we can utilize the Pythagorean Identity, which tells us that cos 2 (10π t ) + sin 2 (10π t ) = 1.
Section 6.5 Modeling with Trigonometric Functions 403 Using this identity, our equation simplifies to 44 − 96 cos(10π t ) + 16 = 0 −96 cos(10π t ) = −60 60 cos(10π t ) = 96 60 cos(u ) = 96 60 u = cos −1 ≈ 0.896 96 u = 2π − 0.896 = 5.388 10π t = 0.896 , so t = 0.0285 10π t = 5.388 , so t = 0.1715
Combine the constants and move to the right side Divide Make a substitution
By symmetry we can find a second solution Undoing the substitution
The point B will be 12 cm from the center of the circle 0.0285 seconds after the process begins, 0.1715 seconds after the process begins, and every 1/5 of a second after each of those values. Important Topics of This Section Modeling with trig equations Modeling with sinusoidal functions Solving right triangles for angles in degrees and radians Try it Now Answers 1. Angle of elevation for the cable is 71.69 degrees and the cable is 73.21 m long π 2. Approximately = G (t ) 66 cos (t − 1) + 87 6
404 Chapter 6
Section 6.5 Exercises In each of the following triangles, solve for the unknown side and angles. 1.
2.
B
B
8
c
3
c
A 5
3.
7
A
B
4. 15
10
7 A
a B
A 12
b
Find a possible formula for the trigonometric function whose values are in the following tables. 5.
6.
x 0 1 2 3 4 5 6 y -2 4 10 4 -2 4 10 x 0 1 2 y 1 -3 -7
3 4 5 -3 1 -3
6 -7
7. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature for the day is 63 degrees and the low temperature of 37 degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t. 8. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature for the day is 92 degrees and the low temperature of 78 degrees occurs at 4 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t. 9. A population of rabbits oscillates 25 above and below an average of 129 during the year, hitting the lowest value in January (t = 0). a. Find an equation for the population, P, in terms of the months since January, t. b. What if the lowest value of the rabbit population occurred in April instead?
Section 6.5 Modeling with Trigonometric Functions 405 10. A population of elk oscillates 150 above and below an average of 720 during the year, hitting the lowest value in January (t = 0). a. Find an equation for the population, P, in terms of the months since January, t. b. What if the lowest value of the rabbit population occurred in March instead? 11. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 105 degrees occurs at 5 PM and the average temperature for the day is 85 degrees. Find the temperature, to the nearest degree, at 9 AM. 12. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 84 degrees occurs at 6 PM and the average temperature for the day is 70 degrees. Find the temperature, to the nearest degree, at 7 AM. 13. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 47 and 63 degrees during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach 51 degrees? 14. Outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 64 and 86 degrees during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach 70 degrees? 15. A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How many minutes of the ride are spent higher than 13 meters above the ground? 16. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? 17. The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square kilometers in March. Assuming sinusoidal fluctuation, during how many months are there less than 9 million square kilometers of sea ice? 18. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming sinusoidal fluctuation, during how many months are there more than 15 million square kilometers of sea ice?
406 Chapter 6 19. A respiratory ailment called “Cheyne-Stokes Respiration” causes the volume per breath to increase and decrease in a sinusoidal manner, as a function of time. For one particular patient with this condition, a machine begins recording a plot of volume per breath versus time (in seconds). Let b(t ) be a function of time t that tells us the volume (in liters) of a breath that starts at time t. During the test, the smallest volume per breath is 0.6 liters and this first occurs for a breath that starts 5 seconds into the test. The largest volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into the test. [UW] a. Find a formula for the function b(t ) whose graph will model the test data for this patient. b. If the patient begins a breath every 5 seconds, what are the breath volumes during the first minute of the test? 20. Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m, at which time the water is 10 feet above the height of low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally. [UW] a. Find a formula for the function y = h(t ) that computes the height of the tide above low tide at time t. (In other words, y = 0 corresponds to low tide.) b. What is the tide height at 11:00 a.m.? 21. A communications satellite orbits the earth t miles above the surface. Assume the radius of the earth is 3,960 miles. The satellite can only “see” a portion of the earth’s surface, bounded by what is called a horizon circle. This leads to a two-dimensional crosssectional picture we can use to study the size of the horizon slice: [UW] a. Find a formula for α in terms of t. b. If t = 30,000 miles, what is α? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? c. If t = 1,000 miles, what is α? What percentage of the circumference of the earth is covered by the satellite? What would be the minimum number of such satellites required to cover the circumference? d. Suppose you wish to place a satellite into orbit so that 20% of the circumference is covered by the satellite. What is the required distance t?
Section 6.5 Modeling with Trigonometric Functions 407 22. Tiffany is a model rocket enthusiast. She has been working on a pressurized rocket filled with nitrous oxide. According to her design, if the atmospheric pressure exerted on the rocket is less than 10 pounds/sq.in., the nitrous oxide chamber inside the rocket will explode. Tiff worked from a formula p = 14.7e − h /10 pounds/sq.in. for the atmospheric pressure h miles above sea level. Assume that the rocket is launched at an angle of α above level ground at sea level with an initial speed of 1400 feet/sec. Also, assume the height (in feet) of the rocket at time t seconds is given by the equation y ( t ) = −16t 2 + 1400sin (α ) t . [UW] a. At what altitude will the rocket explode? b. If the angle of launch is α = 12°, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? c. If the angle of launch is α = 82°, determine the minimum atmospheric pressure exerted on the rocket during its flight. Will the rocket explode in midair? d. Find the largest launch angle α so that the rocket will not explode.
408 Chapter 6
Section 7.1 Solving Trigonometric Equations and Identities 409
Chapter 7: Trigonometric Equations and Identities In the last two chapters we have used basic definitions and relationships to simplify trigonometric expressions and solve trigonometric equations. In this chapter we will look at more complex relationships. By conducting a deeper study of trigonometric identities we can learn to simplify complicated expressions, allowing us to solve more interesting applications. Section 7.1 Solving Trigonometric Equations with Identities .................................... 409 Section 7.2 Addition and Subtraction Identities ......................................................... 417 Section 7.3 Double Angle Identities ........................................................................... 431 Section 7.4 Modeling Changing Amplitude and Midline ........................................... 442
Section 7.1 Solving Trigonometric Equations with Identities
In the last chapter, we solved basic trigonometric equations. In this section, we explore the techniques needed to solve more complicated trig equations. Building from what we already know makes this a much easier task. Consider the function f (= x) 2 x 2 + x . If you were asked to solve f ( x) = 0 , it require simple algebra: Factor 2x 2 + x = 0 Giving solutions x(2 x + 1) = 0 1 x = 0 or x = − 2 Similarly, for g (t ) = sin(t ) , if we asked you to solve g (t ) = 0 , you can solve this using unit circle values: sin(t ) = 0 for t = 0, π , 2π and so on. Using these same concepts, we consider the composition of these two functions: f ( g (t )) = 2(sin(t )) 2 + (sin(t )) = 2 sin 2 (t ) + sin(t ) This creates an equation that is a polynomial trig function. With these types of functions, we use algebraic techniques like factoring and the quadratic formula, along with trigonometric identities and techniques, to solve equations. As a reminder, here are some of the essential trigonometric identities that we have learned so far:
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
410 Chapter 7 Identities Pythagorean Identities cos 2 (t ) + sin 2 (t ) = 1
1 + cot 2 (t ) = csc 2 (t )
1 + tan 2 (t ) = sec 2 (t )
Negative Angle Identities sin(−t ) = − sin(t ) cos(−t ) = cos(t ) csc(−t ) = − csc(t ) sec(−t ) = sec(t ) Reciprocal Identities 1 1 sec(t ) = csc(t ) = cos(t ) sin(t )
tan(t ) =
tan(−t ) = − tan(t ) cot(−t ) = − cot(t )
sin(t ) cos(t )
cot(t ) =
1 tan(t )
Example 1 Solve 2 sin 2 (t ) + sin(t ) = 0 for all solutions with 0 ≤ t < 2π . This equation kind of looks like a quadratic equation, but with sin(t) in place of an algebraic variable (we often call such an equation “quadratic in sine”). As with all quadratic equations, we can use factoring techniques or the quadratic formula. This expression factors nicely, so we proceed by factoring out the common factor of sin(t): sin(t )(2 sin(t ) + 1) = 0 Using the zero product theorem, we know that the product on the left will equal zero if either factor is zero, allowing us to break this equation into two cases: or sin(t ) = 0 2 sin(t ) + 1 = 0 We can solve each of these equations independently From our knowledge of special angles sin(t ) = 0 t = 0 or t = π 2 sin(t ) + 1 = 0 1 sin(t ) = − 2 11π 7π or t = t= 6 6
Again from our knowledge of special angles
Altogether, this gives us four solutions to the equation on 0 ≤ t < 2π : 7π 11π t = 0, π , , 6 6 We could check these answers are reasonable by graphing the function and comparing the zeros.
Section 7.1 Solving Trigonometric Equations and Identities 411 Example 2 Solve 3 sec 2 (t ) − 5 sec(t ) − 2 = 0 for all solutions with 0 ≤ t < 2π . Since the left side of this equation is quadratic in secant, we can try to factor it, and hope it factors nicely. If it is easier to for you to consider factoring without the trig function present, consider using a substitution u = sec(t ) , resulting in 3u 2 − 5u − 2 = 0 , and then try to factor: 3u 2 − 5u − 2 = (3u + 1)(u − 2) Undoing the substitution, (3 sec(t ) + 1)(sec(t ) − 2) = 0 Since we have a product equal to zero, we break it into the two cases and solve each separately. 3 sec(t ) + 1 = 0 1 sec(t ) = − 3 1 1 =− cos(t ) 3 cos(t ) = −3
Isolate the secant Rewrite as a cosine Invert both sides
Since the cosine has a range of [-1, 1], the cosine will never take on an output of -3. There are no solutions to this case. Continuing with the second case, sec(t ) − 2 = 0 sec(t ) = 2 1 =2 cos(t ) 1 cos(t ) = 2
t=
π
3
or t =
Isolate the secant Rewrite as a cosine Invert both sides This gives two solutions 5π 3
These are the only two solutions on the interval. By utilizing technology to graph f (t ) = 3sec 2 (t ) − 5sec(t ) − 2 , a look at a graph confirms there are only two zeros for this function on the interval [0, 2π), which assures us that we didn’t miss anything.
412 Chapter 7 Try it Now 1. Solve 2 sin 2 (t ) + 3 sin(t ) + 1 = 0 for all solutions with 0 ≤ t < 2π . When solving some trigonometric equations, it becomes necessary to first rewrite the equation using trigonometric identities. One of the most common is the Pythagorean Identity, sin 2 (θ ) + cos 2 (θ ) = 1 which allows you to rewrite sin 2 (θ ) in terms of cos 2 (θ ) or vice versa, sin 2 (θ ) = 1 − cos 2 (θ )
cos 2 (θ ) = 1 − sin 2 (θ ) This identity becomes very useful whenever an equation involves a combination of sine and cosine functions. Example 3 Solve 2 sin 2 (t ) − cos(t ) = 1 for all solutions with 0 ≤ t < 2π . Since this equation has a mix of sine and cosine functions, it becomes more complicated to solve. It is usually easier to work with an equation involving only one trig function. This is where we can use the Pythagorean Identity. 2 sin 2 (t ) − cos(t ) = 1 2 1 − cos 2 (t ) − cos(t ) = 1 2 − 2 cos 2 (t ) − cos(t ) = 1
(
)
Using sin 2 (θ ) = 1 − cos 2 (θ ) Distributing the 2
Since this is now quadratic in cosine, we rearrange the equation so one side is zero and factor. Multiply by -1 to simplify the factoring − 2 cos 2 (t ) − cos(t ) + 1 = 0 2 Factor 2 cos (t ) + cos(t ) − 1 = 0 (2 cos(t ) − 1)(cos(t ) + 1) = 0 This product will be zero if either factor is zero, so we can break this into two separate cases and solve each independently. or 2 cos(t ) − 1 = 0 cos(t ) + 1 = 0 1 or cos(t ) = cos(t ) = −1 2 π 5π or t =π t = or t = 3 3
Section 7.1 Solving Trigonometric Equations and Identities 413 Try it Now 2. Solve 2 sin 2 (t ) = 3 cos(t ) for all solutions with 0 ≤ t < 2π . In addition to the Pythagorean Identity, it is often necessary to rewrite the tangent, secant, cosecant, and cotangent as part of solving an equation. Example 4 Solve tan( x) = 3 sin( x) for all solutions with 0 ≤ x < 2π . With a combination of tangent and sine, we might try rewriting tangent tan( x) = 3 sin( x) sin( x) Multiplying both sides by cosine = 3 sin( x) cos( x) sin( x) = 3 sin( x) cos( x)
At this point, you may be tempted to divide both sides of the equation by sin(x). Resist the urge. When we divide both sides of an equation by a quantity, we are assuming the quantity is never zero. In this case, when sin(x) = 0 the equation is satisfied, so we’d lose those solutions if we divided by the sine. To avoid this problem, we can rearrange the equation so that one side is zero 1. Factoring out sin(x) from both parts sin( x) − 3 sin( x) cos( x) = 0 sin( x)(1 − 3 cos( x) ) = 0 From here, we can see we get solutions when sin( x) = 0 or 1 − 3 cos( x) = 0 . Using our knowledge of the special angles of the unit circle sin( x) = 0 when x = 0 or x = π. For the second equation, we will need the inverse cosine. 1 − 3 cos( x) = 0 1 Using our calculator or technology cos( x) = 3 1 Using symmetry to find a second solution x = cos −1 ≈ 1.231 3 x = 2π − 1.231 = 5.052 We have four solutions on 0 ≤ x < 2π : x = 0, 1.231, π, 5.052
1
You technically can divide by sin(x) as long as you separately consider the case where sin(x) = 0. Since it is easy to forget this step, the factoring approach used in the example is recommended.
414 Chapter 7 Try it Now 3. Solve sec(θ ) = 2 cos(θ ) to find the first four positive solutions. Example 5 Solve
4 + 3cos (θ ) = 2 cot (θ ) tan (θ ) for all solutions with 0 ≤ θ < 2π . sec 2 (θ )
4 + 3cos (θ ) = 2 cot (θ ) tan (θ ) sec 2 (θ ) 1 tan(θ ) tan(θ ) 4 cos 2 (θ ) + 3cos (θ ) = 2
4 cos 2 (θ ) + 3 cos(θ ) = 2
Using the reciprocal identities
Simplifying Subtracting 2 from each side
4 cos 2 (θ ) + 3cos (θ ) − 2 = 0 This does not appear to factor nicely so we use the quadratic formula, remembering that we are solving for cos(θ).
− 3 ± 3 2 − 4(4)(−2) − 3 ± 41 cos(θ ) = = 2(4) 8 Using the negative square root first, − 3 − 41 cos(θ ) = = −1.175 8 This has no solutions, since the cosine can’t be less than -1. Using the positive square root, − 3 + 41 cos(θ ) = = 0.425 8 θ = cos −1 (0.425) = 1.131 θ = 2π − 1.131 = 5.152
By symmetry, a second solution can be found
Important Topics of This Section Review of Trig Identities Solving Trig Equations By Factoring Using the Quadratic Formula Utilizing Trig Identities to simplify
Section 7.1 Solving Trigonometric Equations and Identities 415 Try it Now Answers 7π 3π 11π 1. t = , , 6 2 6 π 5π 2. t = , 3 3 π 3π 5π 7π 3. θ = , , , 4 4 4 4
416 Chapter 7
Section 7.1 Exercises Find all solutions on the interval 0 ≤ θ < 2π . 1. 2sin (θ ) = −1 2. 2sin (θ ) = 3 3. 2 cos (θ ) = 1 Find all solutions. π 5. 2sin x =1 4 π 9. 3cos x = 2 5
π 6. 2sin x = 2 3 π 10. 8cos x = 6 2
4. 2 cos (θ ) = 2 −
7. 2 cos ( 2t ) = − 3
8. 2 cos ( 3t ) = −1
11. 7 sin ( 3t ) = −2
12. 4sin ( 4t ) = 1
Find all solutions on the interval [0, 2π ) . 13. 10sin ( x ) cos ( x ) = 6 cos ( x )
14. −3sin ( t ) = 15cos ( t ) sin ( t )
0 15. csc ( 2 x ) − 9 =
16. sec ( 2θ ) = 3
17. sec ( x ) sin ( x ) − 2sin ( x ) = 0
18. tan ( x ) sin ( x ) − sin ( x ) = 0
1 4 2 21. sec x = 7 23. 2sin 2 w + 3sin w + 1 = 0
1 2 2 22. csc t = 3 0 24. 8sin 2 x + 6sin ( x ) + 1 =
19. sin 2 x =
20. cos 2 θ =
25. 2 cos 2 t + cos ( t ) = 1
26. 8cos 2 (θ )= 3 − 2 cos (θ )
15cos ( x ) 27. 4 cos 2 ( x) − 4 =
28. 9sin ( w ) − 2 = 4sin 2 ( w)
0 29. 12sin 2 ( t ) + cos ( t ) − 6 =
0 30. 6 cos 2 ( x ) + 7 sin ( x ) − 8 =
31. cos 2 φ = −6sin φ 33. tan 3 ( x ) = 3 tan ( x )
32. sin 2 t = cos t 34. cos3 ( t ) = − cos ( t )
35. tan 5 ( x ) = tan ( x )
0 36. tan 5 ( x ) − 9 tan ( x ) =
37. 4sin ( x ) cos ( x ) + 2sin ( x ) − 2 cos ( x ) − 1 =0 38. 2sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 =0 0 39. tan ( x ) − 3sin ( x ) =
40. 3cos ( x ) = cot ( x )
41. 2 tan 2 ( t ) = 3sec ( t )
tan 2 ( w ) 42. 1 − 2 tan ( w ) =
Section 7.2 Addition and Subtraction Identities 417
Section 7.2 Addition and Subtraction Identities
In this section, we begin expanding our repertoire of trigonometric identities. Identities The sum and difference identities cos(α − β ) = cos(α ) cos( β ) + sin(α ) sin( β ) cos(α + β ) = cos(α ) cos( β ) − sin(α ) sin( β ) sin(α + β ) = sin(α ) cos( β ) + cos(α ) sin( β ) sin(α − β ) = sin(α ) cos( β ) − cos(α ) sin( β ) We will prove the difference of angles identity for cosine. The rest of the identities can be derived from this one. Proof of the difference of angles identity for cosine Consider two points on a unit circle: P at an angle of α from the positive x axis with coordinates (cos(α ), sin(α ) ) Q at an angle of β with coordinates (cos(β ), sin( β ) ) Notice the measure of angle POQ is α – β. Label two more points: C at an angle of α – β, with coordinates (cos(α − β ), sin(α − β ) ) , D at the point (1, 0).
C P Q α
α-β β
O
Notice that the distance from C to D is the same as the distance from P to Q because triangle COD is a rotation of triangle POQ. Using the distance formula to find the distance from P to Q yields
(cos(α ) − cos(β ) )2 + (sin(α ) − sin( β ) )2 Expanding this cos 2 (α ) − 2 cos(α ) cos( β ) + cos 2 ( β ) + sin 2 (α ) − 2 sin(α ) sin( β ) + sin 2 ( β )
Applying the Pythagorean Identity and simplifying 2 − 2 cos(α ) cos( β ) − 2 sin(α ) sin( β )
D
418 Chapter 7 Similarly, using the distance formula to find the distance from C to D
(cos(α − β ) − 1)2 + (sin(α − β ) − 0)2 Expanding this cos 2 (α − β ) − 2 cos(α − β ) + 1 + sin 2 (α − β )
Applying the Pythagorean Identity and simplifying − 2 cos(α − β ) + 2 Since the two distances are the same we set these two formulas equal to each other and simplify 2 − 2 cos(α ) cos( β ) − 2 sin(α ) sin( β ) = − 2 cos(α − β ) + 2 2 − 2 cos(α ) cos( β ) − 2 sin(α ) sin( β ) = −2 cos(α − β ) + 2 cos(α ) cos( β ) + sin(α ) sin( β ) = cos(α − β ) Establishing the identity. Try it Now 1. By writing cos(α + β ) as cos(α − (− β )) , show the sum of angles identity for cosine follows from the difference of angles identity proven above. The sum and difference of angles identities are often used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles. Example 1 Find the exact value of cos(75°) . Since 75° = 30° + 45° , we can evaluate cos(75°) as Apply the cosine sum of angles identity cos(75°) = cos(30° + 45°) Evaluate = cos(30°) cos(45°) − sin(30°) sin(45°) 3 2 1 2 ⋅ − ⋅ 2 2 2 2 6− 2 = 4
=
Try it Now
π 2. Find the exact value of sin . 12
Simply
Section 7.2 Addition and Subtraction Identities 419 Example 2
π Rewrite sin x − in terms of sin(x) and cos(x). 4
π sin x − 4 π π = sin ( x ) cos − cos( x )sin 4 4 =
Use the difference of angles identity for sine Evaluate the cosine and sine and rearrange
2 2 sin ( x ) − cos( x ) 2 2
Additionally, these identities can be used to simplify expressions or prove new identities Example 3 sin( a + b) tan(a ) + tan(b) Prove . = sin( a − b) tan(a ) − tan(b) As with any identity, we need to first decide which side to begin with. Since the left side involves sum and difference of angles, we might start there sin(a + b) sin( a − b) sin( a ) cos(b) + cos(a ) sin(b) = sin( a ) cos(b) − cos(a ) sin(b)
Apply the sum and difference of angle identities
Since it is not immediately obvious how to proceed, we might start on the other side, and see if the path is more apparent. tan(a ) + tan(b) Rewriting the tangents using the tangent identity tan(a ) − tan(b) sin( a ) sin(b) + cos(a ) cos(b) = sin(a ) sin(b) − cos(a ) cos(b)
Multiplying the top and bottom by cos(a)cos(b)
sin(a ) sin(b) cos(a ) cos(b) + cos(a ) cos(b) Distributing and simplifying = sin(a ) sin(b) cos(a ) cos(b) − cos(a ) cos(b)
420 Chapter 7 =
sin( a ) cos(b) + sin(b) cos(a ) sin( a ) cos(b) − sin(b) cos(a )
From above, we recognize this
=
sin(a + b) sin( a − b)
Establishing the identity
These identities can also be used to solve equations. Example 4 Solve sin( x) sin( 2 x) + cos( x) cos(2 x) =
3 . 2
By recognizing the left side of the equation as the result of the difference of angles identity for cosine, we can simplify the equation 3 Apply the difference of angles identity sin( x) sin( 2 x) + cos( x) cos(2 x) = 2 3 cos( x − 2 x) = 2 3 Use the negative angle identity cos(−x) = 2 3 cos( x) = 2 Since this is a special cosine value we recognize from the unit circle, we can quickly write the answers:
π
+ 2πk 6 , where k is an integer 11π x= + 2πk 6 x=
Combining Waves of Equal Period A sinusoidal function of the form f ( x) = A sin( Bx + C ) can be rewritten using the sum of angles identity. Example 5
π Rewrite f ( x) = 4 sin 3 x + as a sum of sine and cosine. 3
Section 7.2 Addition and Subtraction Identities 421 Using the sum of angles identity π 4 sin 3 x + 3 π π = 4 sin (3x ) cos + cos(3x )sin 3 3 1 3 = 4 sin (3 x ) ⋅ + cos(3 x ) ⋅ 2 2 = 2 sin (3 x ) + 2 3 cos(3 x )
Evaluate the sine and cosine Distribute and simplify
Notice that the result is a stretch of the sine added to a different stretch of the cosine, but both have the same horizontal compression, which results in the same period. We might ask now whether this process can be reversed – can a combination of a sine and cosine of the same period be written as a single sinusoidal function? To explore this, we will look in general at the procedure used in the example above. f ( x) = A sin( Bx + C ) = A(sin( Bx) cos(C ) + cos( Bx) sin(C ) ) = A sin( Bx) cos(C ) + A cos( Bx) sin(C ) = A cos(C ) sin( Bx) + A sin(C ) cos( Bx)
Use the sum of angles identity Distribute the A Rearrange the terms a bit
Based on this result, if we have an expression of the form m sin( Bx) + n cos( Bx) , we could rewrite it as a single sinusoidal function if we can find values A and C so that m sin( Bx) + n cos( Bx) = A cos(C ) sin( Bx) + A sin(C ) cos( Bx) , which will require that: m = cos(C ) m = A cos(C ) which can be rewritten as A n = A sin(C ) n = sin(C ) A To find A, 2 2 m 2 + n 2 = ( A cos(C ) ) + ( A sin(C ) ) = A 2 cos 2 (C ) + A 2 sin 2 (C ) = A 2 cos 2 (C ) + sin 2 (C ) = A2
(
)
Apply the Pythagorean Identity and simplify
Rewriting a Sum of Sine and Cosine as a Single Sine To rewrite m sin( Bx) + n cos( Bx) as A sin( Bx + C ) m n A 2 = m 2 + n 2 , cos(C ) = , and sin(C ) = A A
422 Chapter 7 We can use either of the last two equations to solve for possible values of C. Since there will usually be two possible solutions, we will need to look at both to determine which quadrant C is in and determine which solution for C satisfies both equations. Example 6 Rewrite 4 3 sin(2 x) − 4 cos(2 x) as a single sinusoidal function.
( )
Using the formulas above, A 2 = 4 3 + (− 4 ) = 16 ⋅ 3 + 16 = 64 , so A = 8. Solving for C, 4 3 3 π 11π , so C = or C = . = cos(C ) = 8 2 6 6 −4 1 11π However, since sin(C ) = = − , the angle that works for both is C = 8 2 6 2
2
Combining these results gives us the expression 11π 8 sin 2 x + 6 Try it Now 3. Rewrite − 3 2 sin(5 x) + 3 2 cos(5 x) as a single sinusoidal function. Rewriting a combination of sine and cosine of equal periods as a single sinusoidal function provides an approach for solving some equations. Example 7 Solve 3 sin(2 x) + 4 cos(2 x) = 1 to find two positive solutions. To approach this, since the sine and cosine have the same period, we can rewrite them as a single sinusoidal function. 2 2 A 2 = (3) + (4 ) = 25 , so A = 5 3 3 cos(C ) = , so C = cos −1 ≈ 0.927 or C = 2π − 0.927 = 5.356 5 5 4 Since sin(C ) = , a positive value, we need the angle in the first quadrant, C = 0.927. 5 Using this, our equation becomes Divide by 5 5 sin (2 x + 0.927 ) = 1 1 Make the substitution u = 2x + 0.927 sin (2 x + 0.927 ) = 5
Section 7.2 Addition and Subtraction Identities 423 1 5 1 u = sin −1 ≈ 0.201 5 u = π − 0.201 = 2.940 u = 2π + 0.201 = 6.485
sin (u ) =
The inverse gives a first solution By symmetry, the second solution is A third solution is
Undoing the substitution, we can find two positive solutions for x. 2 x + 0.927 = 0.201 or 2 x + 0.927 = 2.940 or 2 x + 0.927 = 6.485 2 x = −0.726 2 x = 2.013 2 x = 5.558 x = −0.363 x = 1.007 x = 2.779 Since the first of these is negative, we eliminate it and keep the two positive solutions, x = 1.007 and x = 2.779 . The Product-to-Sum and Sum-to-Product Identities Identities The Product-to-Sum Identities 1 sin(α ) cos( β ) = (sin(α + β ) + sin(α − β ) ) 2 1 sin(α ) sin( β ) = (cos(α − β ) − cos(α + β ) ) 2 1 cos(α ) cos( β ) = (cos(α + β ) + cos(α − β ) ) 2 We will prove the first of these, using the sum and difference of angles identities from the beginning of the section. The proofs of the other two identities are similar and are left as an exercise. Proof of the product-to-sum identity for sin(α)cos(β) Recall the sum and difference of angles identities from earlier sin(α + β ) = sin(α ) cos( β ) + cos(α ) sin( β ) sin(α − β ) = sin(α ) cos( β ) − cos(α ) sin( β ) Adding these two equations, we obtain sin(α + β ) + sin(α − β ) = 2 sin(α ) cos( β ) Dividing by 2, we establish the identity 1 sin(α ) cos( β ) = (sin(α + β ) + sin(α − β ) ) 2
424 Chapter 7 Example 8 Write sin(2t ) sin(4t ) as a sum or difference. Using the product-to-sum identity for a product of sines 1 sin( 2t ) sin( 4t ) = (cos(2t − 4t ) − cos(2t + 4t ) ) 2 1 If desired, apply the negative angle identity = (cos(−2t ) − cos(6t ) ) 2 1 Distribute = (cos(2t ) − cos(6t ) ) 2 1 1 = cos(2t ) − cos(6t ) 2 2 Try it Now
11π π 4. Evaluate cos cos . 12 12
Identities The Sum-to-Product Identities u +v u −v sin (u ) + sin (v ) = 2 sin cos 2 2 u −v u +v sin (u ) − sin (v ) = 2 sin cos 2 2 u +v u −v cos(u ) + cos(v ) = 2 cos cos 2 2 u +v u −v cos(u ) − cos(v ) = −2 sin sin 2 2 We will again prove one of these and leave the rest as an exercise. Proof of the sum-to-product identity for sine functions We begin with the product-to-sum identity 1 sin(α ) cos( β ) = (sin(α + β ) + sin(α − β ) ) 2 We define two new variables: u =α +β v =α −β
Section 7.2 Addition and Subtraction Identities 425 u+v 2 u−v Subtracting the equations yields u − v = 2 β , or β = 2
Adding these equations yields u + v = 2α , giving α =
Substituting these expressions into the product-to-sum identity above, u +v u −v 1 Multiply by 2 on both sides sin cos = (sin (u ) + sin (v )) 2 2 2 u +v u −v Establishing the identity 2 sin cos = sin (u ) + sin (v ) 2 2 Example 9 Evaluate cos(15°) − cos(75°) . Using the sum-to-product identity for the difference of cosines, cos(15°) − cos(75°) 15° + 75° 15° − 75° Simplify = −2 sin sin 2 2
= −2 sin (45°)sin (− 30°) = −2 ⋅
Evaluate
2 −1 2 ⋅ = 2 2 2
Example 10 Prove the identity
cos(4t ) − cos(2t ) = − tan(t ) . sin( 4t ) + sin( 2t )
Since the left side seems more complicated, we can start there and simplify. cos(4t ) − cos(2t ) Using the sum-to-product identities sin( 4t ) + sin( 2t ) 4t + 2t 4t − 2t − 2 sin sin 2 2 Simplify = 4t + 2t 4t − 2t 2 sin cos 2 2 − 2 sin (3t )sin (t ) Simplify further = 2 sin (3t ) cos(t ) − sin (t ) Rewrite as a tangent = cos(t ) Establishing the identity = − tan(t )
426 Chapter 7 Try it Now 5. Notice that, using the negative angle identity, sin (u ) − sin (v ) = sin(u ) + sin(−v) . Use this along with the sum of sines identity to prove the sum-to-product identity for sin (u ) − sin (v ) . Example 11 Solve sin (π t ) + sin ( 3π t ) = cos(π t ) for all solutions with 0 ≤ t < 2 . In an equation like this, it is not immediately obvious how to proceed. One option would be to combine the two sine functions on the left side of the equation. Another would be to move the cosine to the left side of the equation, and combine it with one of the sines. For no particularly good reason, we’ll begin by combining the sines on the left side of the equation and see how things work out. Apply the sum to product identity on the left sin (π t ) + sin ( 3π t ) = cos(π t ) π t + 3π t π t − 3π t 2sin cos = cos(π t ) Simplify 2 2 Apply the negative angle identity 2sin ( 2π t ) cos ( −π t ) = cos(π t )
2sin ( 2π t ) cos (π t ) = cos(π t )
Rearrange the equation to be 0 on one side
2sin ( 2π t ) cos (π t ) − cos(π t ) = 0
Factor out the cosine
cos (π t ) ( 2sin ( 2π t ) − 1) = 0
Using the Zero Product Theorem we know that at least one of the two factors must be 2π zero. The first factor, cos (π t ) , has period P = = 2 , so the solution interval of
π
0 ≤ t < 2 represents one full cycle of this function. Substitute u = π t cos (π t ) = 0
cos(u ) = 0 3π π u = or u = 2 2
π
On one cycle, this has solutions Undo the substitution
1 2 2 3π 3 , so t = πt = 2 2
πt =
, so t =
2π = 1 , so the solution interval 2π 0 ≤ t < 2 contains two complete cycles of this function.
The second factor, 2sin ( 2π t ) − 1 , has period of P =
Section 7.2 Addition and Subtraction Identities 427
2sin ( 2π t ) − 1 =0 sin ( 2π t ) = 1 2
sin(u ) = u=
π 6
1 2
π
π 6
u = 2π t
On one cycle, this has solutions
or u =
u = 2π +
=
5π 6
, so t =
Altogether, we found six solutions on 0 ≤ t < 2 , which we can confirm by looking at the graph. 1 5 1 13 3 17 t= , , , , , 12 12 2 12 2 12
Important Topics of This Section The sum and difference identities Combining waves of equal periods Product-to-sum identities Sum-to-product identities Completing proofs Try it Now Answers cos(α + β ) = cos(α − (− β )) 1.
On the second cycle, the solutions are
13π 5π 17π or u = 2π + Undo the substitution = 6 6 6
1 6 12 5π 5 , so t = 2π t = 6 12 13π 13 , so t = 2π t = 12 6 17π 17 , so t = 2π t = 12 6
2π t =
Isolate the sine
cos(α ) cos(− β ) + sin(α ) sin(− β ) cos(α ) cos( β ) + sin(α )(− sin( β )) cos(α ) cos( β ) − sin(α ) sin( β )
428 Chapter 7 6− 2 4 3π 3. 6 sin 5 x + 4 −2− 3 4. 4 5. sin(u ) − sin(v) sin(u ) + sin(−v)
2.
u + (− v ) u − (− v ) 2 sin cos 2 2 u −v u +v 2 sin cos 2 2
Use negative angle identity for sine Use sum-to-product identity for sine Eliminate the parenthesis Establishing the identity
Section 7.2 Addition and Subtraction Identities 429
Section 7.2 Exercises Find an exact value for each of the following. 1. sin ( 75° ) 2. sin (195° ) 3. cos(165°)
4. cos(345°)
5π 7. sin 12
11π 8. sin 12
5π 11. cos x − 6
2π 12. cos x + 3
Simplify each expression. π π 14. sec − w 13. csc − t 2 2
π 15. cot − x 2
π 16. tan − x 2
Rewrite the product as a sum. 17. 16sin (16 x ) sin (11x )
18. 20 cos ( 36t ) cos ( 6t )
19. 2sin ( 5 x ) cos ( 3x )
20. 10 cos ( 5 x ) sin (10 x )
Rewrite the sum as a product. 21. cos ( 6t ) + cos ( 4t )
22. cos ( 6u ) + cos ( 4u )
23. sin ( 3 x ) + sin ( 7 x )
24. sin ( h ) + sin ( 3h )
7π 5. cos 12
π 6. cos 12
Rewrite in terms of sin ( x ) and cos ( x ) . 11π 9. sin x + 6
3π 10. sin x − 4
1 2 and cos ( b ) = − , with a and b both in the interval 4 3 a. Find sin ( a + b ) b. Find cos ( a − b )
25. Given sin ( a ) =
1 4 and cos ( b ) = , with a and b both in the interval 3 5 b. Find cos ( a + b ) a. Find sin ( a − b )
26. Given sin ( a ) =
Solve each equation for all solutions. 0.9 − 27. sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = 28. sin ( 6 x ) cos (11x ) − cos ( 6 x ) sin (11x ) = 0.1 − 29. cos ( 2 x ) cos ( x ) + sin ( 2 x ) sin ( x ) = 1 3 30. cos ( 5 x ) cos ( 3 x ) − sin ( 5 x ) sin ( 3 x ) = 2
π 2 , π :
π 0, 2 :
430 Chapter 7 Solve each equation for all solutions. 31. cos ( 5 x ) = − cos ( 2 x ) 32. sin ( 5 x ) = sin ( 3 x ) 33. cos ( 6θ ) − cos ( 2θ ) = sin ( 4θ ) 34. cos ( 8θ ) − cos ( 2θ ) = sin ( 5θ ) Rewrite as a single function of the form A sin( Bx + C ) . 35. 4sin ( x ) − 6 cos ( x ) 36. − sin ( x ) − 5cos ( x ) 37. 5sin ( 3 x ) + 2 cos ( 3 x )
38. −3sin ( 5 x ) + 4 cos ( 5 x )
Solve for the first two positive solutions. 39. −5sin ( x ) + 3cos ( x ) = 1
40. 3sin ( x ) + cos ( x ) = 2
3 41. 3sin ( 2 x ) − 5cos ( 2 x ) =
1 42. −3sin ( 4 x ) − 2 cos ( 4 x ) =
Simplify. sin ( 7t ) + sin ( 5t ) 43. cos ( 7t ) + cos ( 5t )
44.
Prove the identity. π tan ( x ) + 1 44. tan x + = 4 1 − tan ( x ) π 1 − tan ( t ) 45. tan − t = 4 1 + tan ( t ) 46. cos ( a + b ) + cos ( a − b ) = 2 cos ( a ) cos ( b ) 47. 48.
cos ( a + b ) cos ( a − b ) tan ( a + b ) tan ( a − b )
= =
1 − tan ( a ) tan ( b )
1 + tan ( a ) tan ( b )
sin ( a ) cos ( a ) + sin ( b ) cos ( b ) sin ( a ) cos ( a ) − sin ( b ) cos ( b )
= b ) cos ( 2b ) − cos(2a) 49. 2sin ( a + b ) sin ( a − sin ( x ) + sin ( y ) 1 = tan ( x + y ) cos ( x ) + cos ( y ) 2 cos ( a + b ) 51. = 1 − tan ( a ) tan ( b ) cos ( a ) cos ( b )
50.
52. cos ( x + y ) cos ( x − y= ) cos 2 x − sin 2 y
sin ( 9t ) − sin ( 3t )
cos ( 9t ) + cos ( 3t )
Section 7.3 Double Angle Identities 431
Section 7.3 Double Angle Identities
Two special cases of the sum of angles identities arise often enough that we choose to state these identities separately. Identities The double angle identities sin(2α ) = 2 sin(α ) cos(α ) cos(2α ) = cos 2 (α ) − sin 2 (α ) = 1 − 2 sin 2 (α ) =
2 cos 2 (α ) − 1
These identities follow from the sum of angles identities. Proof of the sine double angle identity sin(2α ) Apply the sum of angles identity = sin(α + α ) Simplify = sin(α ) cos(α ) + cos(α ) sin(α ) Establishing the identity = 2 sin(α ) cos(α ) Try it Now 1. Show cos(2α ) = cos 2 (α ) − sin 2 (α ) by using the sum of angles identity for cosine. For the cosine double angle identity, there are three forms of the identity stated because the basic form, cos(2α ) = cos 2 (α ) − sin 2 (α ) , can be rewritten using the Pythagorean Identity. Rearranging the Pythagorean Identity results in the equality cos 2 (α ) = 1 − sin 2 (α ) , and by substituting this into the basic double angle identity, we obtain the second form of the double angle identity. cos(2α ) = cos 2 (α ) − sin 2 (α ) cos(2α ) = 1 − sin 2 (α ) − sin 2 (α ) cos(2α ) = 1 − 2 sin 2 (α )
Substituting using the Pythagorean identity Simplifying
432 Chapter 7 Example 1 If sin(θ ) = cos(2θ ) .
3 and θ is in the second quadrant, find exact values for sin(2θ ) and 5
To evaluate cos(2θ ) , since we know the value for sin(θ ) , we can use the version of the double angle that only involves sine. 2
18 7 3 cos(2θ ) = 1 − 2 sin 2 (θ ) = 1 − 2 = 1 − = 25 25 5
Since the double angle for sine involves both sine and cosine, we’ll need to first find cos(θ ) , which we can do using the Pythagorean Identity. sin 2 (θ ) + cos 2 (θ ) = 1 2
3 2 + cos (θ ) = 1 5 9 cos 2 (θ ) = 1 − 25 16 4 cos(θ ) = ± =± 25 5
Since θ is in the second quadrant, we know that cos(θ) < 0, so 4 cos(θ ) = − 5 Now we can evaluate the sine double angle 24 3 4 sin( 2θ ) = 2 sin(θ ) cos(θ ) = 2 − = − 25 5 5 Example 2 Simplify the expressions a) 2 cos 2 (12°) − 1
b) 8 sin (3 x ) cos(3 x )
a) Notice that the expression is in the same form as one version of the double angle identity for cosine: cos(2θ ) = 2 cos 2 (θ ) − 1 . Using this, 2 cos 2 (12°) − 1 = cos(2 ⋅ 12°) = cos(24°)
b) This expression looks similar to the result of the double angle identity for sine. Factoring a 4 out of the original expression 8 sin (3 x ) cos(3 x ) Applying the double angle identity 4 ⋅ 2 sin (3 x ) cos(3 x ) 4 sin(6 x)
Section 7.3 Double Angle Identities 433 We can use the double angle identities to simplify expressions and prove identities. Example 2 Simplify
cos(2t ) . cos(t ) − sin(t )
With three choices for how to rewrite the double angle, we need to consider which will be the most useful. To simplify this expression, it would be great if the denominator would cancel with something in the numerator, which would require a factor of cos(t ) − sin(t ) in the numerator, which is most likely to occur if we rewrite the numerator with a mix of sine and cosine. cos(2t ) cos(t ) − sin(t )
Apply the double angle identity
cos 2 (t ) − sin 2 (t ) cos(t ) − sin(t ) (cos(t ) − sin(t ) )(cos(t ) + sin(t ) ) = cos(t ) − sin(t ) = cos(t ) + sin(t )
=
Factor the numerator Cancelling the common factor Resulting in the most simplified form
Example 3 Prove sec(2α ) =
sec 2 (α ) . 2 − sec 2 (α )
Since the right side seems a bit more complicated than the left side, we begin there. sec 2 (α ) Rewrite the secants in terms of cosine 2 − sec 2 (α ) 1 cos 2 (α ) = 1 2− cos 2 (α ) At this point, we could rewrite the bottom with common denominators, subtract the terms, invert and multiply, then simplify. Alternatively, we can multiple both the top and bottom by cos 2 (α ) , the common denominator: 1 ⋅ cos 2 (α ) cos 2 (α ) Distribute on the bottom = 1 2 2 − ⋅ cos (α ) cos 2 (α )
434 Chapter 7 cos 2 (α ) cos 2 (α ) = cos 2 (α ) 2 cos 2 (α ) − ⋅ cos 2 (α ) 1 = 2 2 cos (α ) − 1 1 = cos(2α ) = sec(2α )
Simplify
Rewrite the denominator as a double angle Rewrite as a secant Establishing the identity
Try it Now 2. Use an identity to find the exact value of cos 2 (75°) − sin 2 (75°) . As with other identities, we can also use the double angle identities for solving equations. Example 4 Solve cos(2t ) = cos(t ) for all solutions with 0 ≤ t < 2π . In general when solving trig equations, it makes things more complicated when we have a mix of sines and cosines and when we have a mix of functions with different periods. In this case, we can use a double angle identity to rewrite the cos(2t). When choosing which form of the double angle identity to use, we notice that we have a cosine on the right side of the equation. We try to limit our equation to one trig function, which we can do by choosing the version of the double angle formula for cosine that only involves cosine. cos(2t ) = cos(t )
Apply the double angle identity This is quadratic in cosine, so make one side 0 Factor Break this apart to solve each part separately
2
2 cos (t ) − 1 = cos(t ) 2 cos 2 (t ) − cos(t ) − 1 = 0 (2 cos(t ) + 1)(cos(t ) − 1) = 0 2 cos(t ) + 1 = 0 1 cos(t ) = − 2 4π 2π or t = t= 3 3
or
cos(t ) − 1 = 0
or
cos(t ) = 1
or
t=0
Looking at a graph of cos(2t) and cos(t) shown together, we can verify that these three solutions on [0, 2π) seem reasonable.
Section 7.3 Double Angle Identities 435 Example 5 A cannonball is fired with velocity of 100 meters per second. If it is launched at an angle of θ, the vertical component of the velocity will be 100 sin(θ ) and the horizontal component will be 100 cos(θ ) . Ignoring wind resistance, the height of the cannonball will follow the equation h(t ) = −4.9t 2 + 100 sin(θ )t and horizontal position will follow the equation x(t ) = 100 cos(θ )t . If you want to hit a target 900 meters away, at what angle should you aim the cannon? To hit the target 900 meters away, we want x(t ) = 900 at the time when the cannonball hits the ground, when h(t ) = 0 . To solve this problem, we will first solve for the time, t, when the cannonball hits the ground. Our answer will depend upon the angle θ . h(t ) = 0
− 4.9t 2 + 100 sin(θ )t = 0 t (− 4.9t + 100 sin(θ ) ) = 0 t = 0 or − 4.9t + 100 sin(θ ) = 0 − 4.9t = −100 sin(θ ) 100 sin(θ ) t= 4.9
Factor Break this apart to find two solutions Solve for t
This shows that the height is 0 twice, once at t = 0 when the cannonball is fired, and again when the cannonball hits the ground after flying through the air. This second value of t gives the time when the ball hits the ground in terms of the angle θ . We want the horizontal distance x(t) to be 900 when the ball hits the ground, in other words 100 sin(θ ) when t = . 4.9 Since the target is 900 m away we start with x(t ) = 900
Use the formula for x(t)
100 cos(θ )t = 900 100 sin(θ ) 100 cos(θ ) = 900 4.9 100 2 cos(θ ) sin(θ ) = 900 4.9 900(4.9) cos(θ ) sin(θ ) = 100 2
Substitute the desired time, t from above Simplify Isolate the cosine and sine product
436 Chapter 7 The left side of this equation almost looks like the result of the double angle identity for sine: sin(2θ ) = 2 sin (θ ) cos(θ ) . By dividing both sides of the double angle identity by 2, we get 1 sin( 2α ) = sin(α ) cos(α ) . Applying this to the equation above, 2 1 900(4.9) Multiply by 2 sin( 2θ ) = 2 100 2 2(900)(4.9) Use the inverse sine sin( 2θ ) = 100 2 2(900)(4.9) Divide by 2 2θ = sin −1 ≈ 1.080 2 100 1.080 θ= = 0.540 , or about 30.94 degrees 2
Power Reduction and Half Angle Identities Another use of the cosine double angle identities is to use them in reverse to rewrite a squared sine or cosine in terms of the double angle. Starting with one form of the cosine double angle identity: Isolate the cosine squared term cos(2α ) = 2 cos 2 (α ) − 1 2 Add 1 cos(2α ) + 1 = 2 cos (α ) cos(2α ) + 1 Divide by 2 cos 2 (α ) = 2 cos(2α ) + 1 This is called a power reduction identity cos 2 (α ) = 2 Try it Now 3. Use another form of the cosine double angle identity to prove the identity 1 − cos(2α ) . sin 2 (α ) = 2 Example 6 Rewrite cos 4 ( x) without any powers.
(
)
2
Since cos 4 ( x) = cos 2 ( x) , we can use the formula we found above
(
cos 4 ( x) = cos 2 ( x)
)
2
Section 7.3 Double Angle Identities 437 cos(2 x) + 1 = 2
( cos(2 x) + 1) =
2
Square the numerator and denominator
2
4 2 cos (2 x) + 2 cos(2 x) + 1 = 4 2 cos (2 x) 2 cos(2 x) 1 = + + 4 4 4
Expand the numerator Split apart the fraction Apply the formula above to cos 2 (2 x) cos 2 (2 x) =
cos(4 x) + 1 2 + 2 cos(2 x) + 1 = 4 4 4 1 cos(4 x) 1 1 = + + cos(2 x) + 8 8 2 4 cos(4 x) 1 3 = + cos(2 x) + 8 2 8
cos(2 ⋅ 2 x) + 1 2
Simplify Combine the constants
The cosine double angle identities can also be used in reverse for evaluating angles that cos(2α ) + 1 are half of a common angle. Building from our formula cos 2 (α ) = , if we let 2 θ θ cos(θ ) + 1 . Taking the square θ = 2α , then α = this identity becomes cos 2 = 2 2 2 root, we obtain cos(θ ) + 1 θ , where the sign is determined by the quadrant. cos = ± 2 2 This is called a half-angle identity. Try it Now 4. Use your results from the last Try it Now to prove the identity 1 − cos(θ ) θ . sin = ± 2 2 Example 7 Find an exact value for cos(15°) . Since 15 degrees is half of 30 degrees, we can use our result from above:
438 Chapter 7 cos(30°) + 1 30° cos(15°) = cos =± 2 2 We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the positive result. cos(30°) + 1 = 2 =
3 +1 2 2
3 1 + 4 2
Identities Half-Angle Identities cos(θ ) + 1 θ cos = ± 2 2 Power Reduction Identities cos(2α ) + 1 cos 2 (α ) = 2
1 − cos(θ ) θ sin = ± 2 2
sin 2 (α ) =
1 − cos(2α ) 2
Since these identities are easy to derive from the double-angle identities, the power reduction and half-angle identities are not ones you should need to memorize separately. Important Topics of This Section Double angle identity Power reduction identity Half angle identity Using identities Simplify equations Prove identities Solve equations Try it Now Answers cos(2α ) = cos(α + α ) 1. cos(α ) cos(α ) − sin(α ) sin(α ) cos 2 (α ) − sin 2 (α )
Section 7.3 Double Angle Identities 439
2. cos(150°) =
− 3 2
1 − cos(2α ) 2 1 − ( cos 2 (α ) − sin 2 (α ) ) 2 1 − cos (α ) + sin 2 (α ) 3. 2 2 sin (α ) + sin 2 (α ) 2 2 2sin (α ) = sin 2 (α ) 2 2
1 − cos(2α ) 2 1 − cos(2α ) sin(α ) = ± 2
sin 2 (α ) =
4. α =
θ
2
θ 1 − cos 2 θ 2 sin = ± 2 2 1 − cos(θ ) θ sin = ± 2 2
440 Chapter 7
Section 7.3 Exercises 1 and x is in quadrant I, then find exact values for (without solving for x): 8 b. cos ( 2x ) c. tan ( 2x ) a. sin ( 2x )
1. If sin ( x ) =
2 and x is in quadrant I, then find exact values for (without solving for x): 3 b. cos ( 2x ) c. tan ( 2x ) a. sin ( 2x )
2. If cos ( x ) =
Simplify each expression. 3. cos 2 ( 28° ) − sin 2 (28°)
4. 2 cos 2 ( 37° ) − 1
5. 1 − 2sin 2 (17°)
6. cos 2 ( 37° ) − sin 2 (37°)
7. cos 2 ( 9 x ) − sin 2 (9 x)
8. cos 2 ( 6 x ) − sin 2 (6 x)
9. 4sin ( 8 x ) cos(8 x)
10. 6sin ( 5 x ) cos(5 x)
Solve for all solutions on the interval [0, 2π ) . 11. 6sin ( 2t ) + 9sin ( t ) = 12. 2sin ( 2t ) + 3cos ( t ) = 0 0 13. 9= cos ( 2θ ) 9 cos 2 (θ ) − 4
14. 8cos = ( 2α ) 8cos 2 (α ) − 1
15. sin ( 2t ) = cos ( t )
16. cos ( 2t ) = sin ( t )
17. cos ( 6 x ) − cos ( 3 x ) = 0
18. sin ( 4 x ) − sin ( 2 x ) = 0
Use a double angle, half angle, or power reduction formula to rewrite without exponents. 19. cos 2 (5 x) 20. cos 2 (6 x) 21. sin 4 (8 x)
22. sin 4 ( 3x )
23. cos 2 x sin 4 x
24. cos 4 x sin 2 x
25. If csc ( x ) = 7 and 90° < x < 180° , then find exact values for (without solving for x): x a. sin 2
x b. cos 2
x c. tan 2
26. If sec ( x ) = 4 and 90° < x < 180° , then find exact values for (without solving for x): x a. sin 2
x b. cos 2
x c. tan 2
Section 7.3 Double Angle Identities 441 Prove the identity. 2 27. ( sin t − cos t ) = 1 − sin ( 2t ) 28. ( sin 2 x −= 1) cos ( 2 x ) + sin 4 x 2
29. sin ( 2 x ) =
2 tan ( x )
1 + tan 2 ( x )
30. tan ( 2 x ) =
2sin ( x ) cos ( x ) 2 cos 2 ( x ) − 1
31. cot ( x ) − tan ( x ) = 2 cot ( 2 x ) 32.
sin ( 2θ )
1 + cos ( 2θ )
33. cos ( 2α ) = 34.
= tan (θ )
1 − tan 2 (α )
1 + tan 2 (α )
1 + cos ( 2t )
sin ( 2t ) − cos ( t )
=
2 cos ( t )
2sin ( t ) − 1
35. sin ( 3 x ) 3sin ( x ) cos 2 ( x ) − sin 3 ( x) = = 36. cos ( 3x ) cos3 ( x) − 3sin 2 ( x) cos ( x )
442 Chapter 7
Section 7.4 Modeling Changing Amplitude and Midline
While sinusoidal functions can model a variety of behaviors, it is often necessary to combine sinusoidal functions with linear and exponential curves to model real applications and behaviors. We begin this section by looking at changes to the midline of a sinusoidal function. Recall that the midline describes the middle, or average value, of the sinusoidal function. Changing Midlines Example 1 A population of elk currently averages 2000 elk, and that average has been growing by 4% each year. Due to seasonal fluctuation, the population oscillates from 50 below average in the winter up to 50 above average in the summer. Find a function that models the number of elk after t years. There are two components to the behavior of the elk population: the changing average, and the oscillation. The average is an exponential growth, starting at 2000 and growing by 4% each year. Writing a formula for this: average = initial (1 + r= )t 2000(1 + 0.04)t For the oscillation, since the population oscillates 50 above and below average, the amplitude will be 50. Since it takes one year for the population to cycle, the period is 1. original period 2π We find the value of the horizontal stretch coefficient= B = = 2π . new period 1 Additionally, since we weren’t told when t was first measured we will have to decide if t = 0 corresponds to winter, or summer. If we choose winter then the shape of the function would be a negative cosine, since it starts at the lowest value. Putting it all together, the equation would be: P (t ) = −50 cos(2π t ) + midline Since the midline represents the average population, we substitute in the exponential function into the population equation to find our final equation: P(t ) = −50 cos(2π t ) + 2000(1 + 0.04)t This is an example of changing midline – in this case an exponentially changing midline. Changing Midline A function of the form f (t ) = A sin( Bt ) + g (t ) will oscillate above and below the average given by the function g(t).
Section 7.4 Modeling Changing Amplitude and Midline 443 Changing midlines can be exponential, linear, or any other type of function. Here are some examples: Linear midline
Exponential midline
= f (t ) A sin ( Bt ) + (mt + b= ) f (t ) A sin ( Bt ) + (abt )
Quadratic midline
= f (t ) A sin ( Bt ) + (at 2 )
Example 2
π Find a function with linear midline of the form f (t ) = A sin t + mt + b that will pass 2 through the points given below.
t f(t)
0 5
1 10
2 9
3 8
Since we are given the value of the horizontal compression coefficient we can calculate original period 2π the period of this function: new period = = = 4. π B 2 Since the sine function is at the midline at the beginning of a cycle and halfway through a cycle, we would expect this function to be at the midline at t = 0 and t = 2, since 2 is half the full period of 4. Based on this, we expect the points (0, 5) and (2, 9) to be points on the midline. We can clearly see that this is not a constant function and so we use the two points to calculate a linear function: midline = mt + b . From these two points we can calculate a slope: 9−5 4 m= = =2 2−0 2 Combining this with the initial value of 5, we have the midline: midline = 2t + 5 , giving π a full function of the form f (t ) = A sin t + 2t + 5 . To find the amplitude, we can 2 plug in a point we haven’t already used, such as (1, 10) π Evaluate the sine and combine like terms 10 = A sin (1) + 2(1) + 5 2 10 = A + 7 A=3
444 Chapter 7 A function of the form given fitting the data would be π f (t ) = 3 sin t + 2t + 5 2 Alternative Approach Notice we could have taken an alternate approach by plugging points (0, 5) and (2, 9) into the original equation. Substituting (0, 5), π Evaluate the sine and simplify 5 = A sin (0) + m(0) + b 2 5=b Substituting (2, 9) π Evaluate the sine and simplify 9 = A sin (2) + m(2) + 5 2 9 = 2m + 5 4 = 2m m = 2 , as we found above. Now we can proceed to find A the same way we did before. Example 3 The number of tourists visiting a ski and hiking resort averages 4000 people annually and oscillates seasonally, 1000 above and below the average. Due to a marketing campaign, the average number of tourists has been increasing by 200 each year. Write an equation for the number of tourists after t years, beginning at the peak season. Again there are two components to this problem: the oscillation and the average. For the oscillation, the number of tourists oscillates 1000 above and below average, giving an amplitude of 1000. Since the oscillation is seasonal, it has a period of 1 year. Since we are given a starting point of “peak season”, we will model this scenario with a cosine function. So far, this gives an equation in the form N (t ) 1000 cos(2π t ) + midline = The average is currently 4000, and is increasing by 200 each year. This is a constant rate of change, so this is linear growth, average = 4000 + 200t . Combining these two pieces gives a function for the number of tourists: = N (t ) 1000 cos(2π t ) + 4000 + 200t Try it Now 1. Given the function g ( x) = ( x 2 − 1) + 8cos( x) , describe the midline and amplitude using words.
Section 7.4 Modeling Changing Amplitude and Midline 445 Changing Amplitude There are also situations in which the amplitude of a sinusoidal function does not stay constant. Back in Chapter 6, we modeled the motion of a spring using a sinusoidal function, but had to ignore friction in doing so. If there were friction in the system, we would expect the amplitude of the oscillation to decrease over time. In the equation f (t ) = A sin( Bt ) + k , A gives the amplitude of the oscillation, we can allow the amplitude to change by replacing this constant A with a function A(t). Changing Amplitude A function of the form f (t ) = A(t ) sin( Bt ) + k will oscillate above and below the midline with an amplitude given by A(t). When thinking about a spring with amplitude decreasing over time, it is tempting to use the simplest tool for the job – a linear function. But if we attempt to model the amplitude with a decreasing linear function, such as A(t ) = 10 − t , we quickly see the problem when we graph the equation f (t ) = (10 − t ) sin(4t ) .
While the amplitude decreases at first as intended, the amplitude hits zero at t = 10, then continues past the intercept, increasing in absolute value, which is not the expected behavior. This behavior and function may model the situation on a restricted domain and we might try to chalk the rest of it up to model breakdown, but in fact springs just don’t behave like this. A better model, as you will learn later in physics and calculus, would show the amplitude decreasing by a fixed percentage each second, leading to an exponential decay model for the amplitude. Damped Harmonic Motion Damped harmonic motion, exhibited by springs subject to friction, follows a model of the form f (t ) = ab t sin( Bt ) + k or f (t ) = ae rt sin( Bt ) + k .
446 Chapter 7 Example 4 A spring with natural length of 20 inches is pulled back 6 inches and released. It oscillates once every 2 seconds. Its amplitude decreases by 20% each second. Find a function that models the position of the spring t seconds after being released. Since the spring will oscillate on either side of the natural length, the midline will be at 20 inches. The oscillation has a period of 2 seconds, and so the horizontal compression coefficient is B = π . Additionally, it begins at the furthest distance from the wall, indicating a cosine model. Meanwhile, the amplitude begins at 6 inches, and decreases by 20% each second, giving an amplitude function of A(t ) = 6(1 − 0.20) t . Combining this with the sinusoidal information gives a function for the position of the spring: f (t ) = 6(0.80) t cos(π t ) + 20 Example 5 A spring with natural length of 30 cm is pulled out 10 cm and released. It oscillates 4 times per second. After 2 seconds, the amplitude has decreased to 5 cm. Find a function that models the position of the spring.
2π 1 second, so = B = 8π . Since the spring will 1 4 4 oscillate on either side of the natural length, the midline will be at 30 cm. It begins at the furthest distance from the wall, suggesting a cosine model. Together, this gives = f (t ) A(t ) cos(8π t ) + 30 . The oscillation has a period of
For the amplitude function, we notice that the amplitude starts at 10 cm, and decreases to 5 cm after 2 seconds. This gives two points (0, 10) and (2, 5) that must be satisfied by an exponential function: A(0) = 10 and A(2) = 5 . Since the function is exponential, we can use the form A(t ) = ab t . Substituting the first point, 10 = ab 0 , so a = 10. Substituting in the second point, Divide by 10 5 = 10b 2 1 Take the square root = b2 2 1 b= ≈ 0.707 2
Section 7.4 Modeling Changing Amplitude and Midline 447 This gives an amplitude function of A(t ) = 10(0.707) t . Combining this with the oscillation, = f (t ) 10(0.707)t cos(8π t ) + 30 Try it Now 2. A certain stock started at a high value of $7 per share and has been oscillating above and below the average value, with the oscillation decreasing by 2% per year. However, the average value started at $4 per share and has grown linearly by 50 cents per year. a. Find a formula for the midline b. Find a formula for the amplitude. c. Find a function S(t) that models the value of the stock after t years. Example 6 In AM (Amplitude Modulated) radio, a carrier wave with a high frequency is used to transmit music or other signals by applying the to-be-transmitted signal as the amplitude of the carrier signal. A musical note with frequency 110 Hz (Hertz = cycles per second) is to be carried on a wave with frequency of 2 KHz (KiloHertz = thousands of cycles per second). If the musical wave has an amplitude of 3, write a function describing the broadcast wave. 1 2000 of a second, giving an equation of the form sin(4000π t ) . Our choice of a sine function here was arbitrary – it would have worked just was well to use a cosine.
The carrier wave, with a frequency of 2000 cycles per second, would have period
The musical tone, with a frequency of 110 cycles per second, would have a period of 1 of a second. With an amplitude of 3, this would correspond to a function of the 110 form 3sin(220π t ) . Again our choice of using a sine function is arbitrary. The musical wave is acting as the amplitude of the carrier wave, so we will multiply the musical tone’s function by the carrier wave function, resulting in the function f (t ) = 3sin(220π t ) sin(4000π t )
448 Chapter 7 Important Topics of This Section Changing midline Changing amplitude Linear Changes Exponential Changes Damped Harmonic Motion Try it Now Answers 1. The midline follows the path of the quadratic x 2 − 1 and the amplitude is a constant value of 8. 2.
m(t )= 4 + 0.5t A(t ) = 7(0.98)t π S(t)= 7(0.98)t cos t + 4 + 0.5t 6
Section 7.4 Modeling Changing Amplitude and Midline 449
Section 7.4 Exercises Find a possible formula for the trigonometric function whose values are given in the following tables. 1.
x 0 3 6 9 12 15 18 y -4 -1 2 -1 -4 -1 2
2.
x 0 2 4 6 8 10 12 y 5 1 -3 1 5 1 -3
3. The displacement h(t ) , in centimeters, of a mass suspended by a spring is modeled by the function h ( t ) = 8sin(6π t ) , where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. 4. The displacement h(t ) , in centimeters, of a mass suspended by a spring is modeled by the function h ( t ) = 11sin(12π t ) , where t is measured in seconds. Find the amplitude, period, and frequency of this displacement.
5. A population of rabbits oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 160 each year. Find a function that models the population, P, in terms of the months since January, t. 6. A population of deer oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 110 each year. Find a function that models the population, P, in terms of the months since January, t. 7. A population of muskrats oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, P, in terms of the months since January, t. 8. A population of fish oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, P, in terms of the months since January, t. 9. A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, D, the end of the spring is below equilibrium in terms of seconds, t, since the spring was released.
450 Chapter 7 10. A spring is attached to the ceiling and pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, D, the end of the spring is below equilibrium in terms of seconds, t, since the spring was released. 11. A spring is attached to the ceiling and pulled 17 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, D the end of the spring is below equilibrium in terms of seconds, t, since the spring was released. 12. A spring is attached to the ceiling and pulled 19 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, D the end of the spring is below equilibrium in terms of seconds, t, since the spring was released. Match each equation form with one of the graphs. 13. a. ab x + sin ( 5 x ) b. sin ( 5x ) + mx + b 14. a. ab x sin ( 5 x )
I
b. ( mx + b ) sin(5 x)
II
III
π Find a function of the form= y ab x + c sin 2 15. x 0 1 2 3 y 6 29 96 379 π Find a function of the = form y a sin 2 17. x 0 1 2 3 y 7 6 11 16
IV x that fits the data given. 16. x 0 1 2 3 y 6 34 150 746
x + m + bx that fits the data given. 18. x 0 1 2 3 y -2 6 4 2
π Find a function of the form y = ab x cos 2 19. x 0 1 2 3 y 11 3 1 3
x + c that fits the data given. 20. x 0 1 2 3 y 4 1 -11 1
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 451
Chapter 8: Further Applications of Trigonometry
In this chapter, we will explore additional applications of trigonometry. We will begin with an extension of the right triangle trigonometry we explored in Chapter 5 to situations involving non-right triangles. We will explore the polar coordinate system and parametric equations as new ways of describing curves in the plane. In the process, we will introduce vectors and an alternative way of writing complex numbers, two important mathematical tools we use when analyzing and modeling the world around us. Section 8.1 Non-right Triangles: Law of Sines and Cosines ...................................... 451 Section 8.2 Polar Coordinates ..................................................................................... 467 Section 8.3 Polar Form of Complex Numbers ............................................................ 480 Section 8.4 Vectors ..................................................................................................... 491 Section 8.5 Parametric Equations ............................................................................... 504
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines
Although right triangles allow us to solve many applications, it is more common to find scenarios where the triangle we are interested in does not have a right angle. Two radar stations located 20 miles apart both detect a UFO located between them. The angle of elevation measured by the first station is 35 degrees. The angle of elevation measured by the second station is 15 degrees. What is the altitude of the UFO?
15°
20 miles
35°
We see that the triangle formed by the UFO and the two stations is not a right triangle. Of course, in any triangle we could draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles, but it would be nice to have methods for working directly with non-right triangles. In this section we will expand upon the right triangle trigonometry we learned in Chapter 5, and adapt it to non-right triangles. Law of Sines Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label h, to create two right triangles. Using the right triangle relationships, h h sin(α ) = and sin( β ) = . b a
b
a h
α
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011. This material is licensed under a Creative Commons CC-BY-SA license.
β
452 Chapter 8 Solving both equations for h, we get b sin(α ) = h and a sin( β ) = h . Since the h is the same in both equations, we establish b sin(α ) = a sin( β ) . Dividing, we conclude that sin(α ) sin( β ) = a b Had we drawn the altitude to be perpendicular to side b or a, we could similarly establish sin(α ) sin(γ ) sin( β ) sin(γ ) and = = a c b c Collectively, these relationships are called the Law of Sines. Law of Sines Given a triangle with angles and sides opposite labeled as shown, the ratio of sine of angle to length of the opposite side will always be equal, or, symbolically, sin(α ) sin( β ) sin(γ ) = = γ a b c a b For clarity, we call side a the corresponding side of angle α. Similarly, we call angle α, the corresponding angle of side a. α β Likewise for side b and angle β, and for side c and angle γ. c When we use the law of sines, we use any pair of ratios as an equation. In the most straightforward case, we know two angles and one of the corresponding sides. Example 1 In the triangle shown here, solve for the unknown sides and angle. Solving for the unknown angle is relatively easy, since the three angles must add to 180 degrees. From this, we can determine that γ = 180° – 50° – 30° = 100°.
b 50°
γ
10 c
30°
To find an unknown side, we need to know the corresponding angle, and we also need another known ratio. Since we know the angle 50° and its corresponding side, we can use this for one of the two ratios. To look for side b, we would use its corresponding angle, 30°.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 453 sin(50°) sin(30°) = 10 b sin(50°) b = sin(30°) 10 10 b = sin(30°) ≈ 6.527 sin(50°)
Multiply both sides by b Divide, or multiply by the reciprocal, to solve for b
Similarly, to solve for side c, we set up the equation sin(50°) sin(100°) = 10 c 10 c = sin(100°) ≈ 12.856 sin(50°) Example 2 Find the elevation of the UFO from the beginning of the section. To find the elevation of the UFO, we first find the distance from one station to the UFO, such as the side a in the picture, then use right triangle relationships to find the height of the UFO, h.
a 15°
h 20 miles
35°
Since the angles in the triangle add to 180 degrees, the unknown angle of the triangle must be 180° – 15° – 35° = 130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship: sin(130°) sin(35°) Multiply by a = 20 a sin(130°) Divide, or multiply by the reciprocal, to solve for a a = sin(35°) 20 20 sin(35°) Simplify a= ≈ 14.975 sin(130°) The distance from one station to the UFO is about 15 miles. Now that we know a, we can use right triangle relationships to solve for h. opposite h h Solve for h sin(15°) = = = hypotenuse a 14.975 h = 14.975 sin(15°) ≈ 3.876
The UFO is at an altitude of 3.876 miles.
454 Chapter 8 In addition to solving triangles in which two angles are known, the law of sines can be used to solve for an angle when two sides and one corresponding angle are known. Example 3 In the triangle shown here, solve for the unknown sides and angles.
12
β
In choosing which pair of ratios from the Law of Sines to α a use, we always want to pick a pair where we know three of the four pieces of information in the equation. In this case, 85° 9 we know the angle 85° and its corresponding side, so we will use that ratio. Since our only other known information is the side with length 9, we will use that side and solve for its corresponding angle. sin(85°) sin( β ) Isolate the unknown = 12 9 9 sin(85°) Use the inverse sine to find a first solution = sin( β ) 12 Remember when we use the inverse function that there are two possible answers. 9 sin(85°) ≈ 48.3438° By symmetry we find the second possible solution 12 β = 180° − 48.3438° = 131.6562°
β = sin −1
In this second case, if β ≈ 132°, then α would be α = 180° – 85° – 132° = –37°, which doesn’t make sense, so the only possibility for this triangle is β = 48.3438°. With a second angle, we can now easily find the third angle, since the angles must add to 180°, so α = 180° – 85° – 48.3438° = 46.6562°. Now that we know α, we can proceed as in earlier examples to find the unknown side a. sin(85°) sin( 46.6562°) = 12 a 12 sin( 46.6562°) a= ≈ 8.7603 sin(85°) Notice that in the problem above, when we use Law of Sines to solve for an unknown angle, there can be two possible solutions. This is called the ambiguous case, and can arise when we know two sides and a non-included angle. In the ambiguous case we may find that a particular set of given information can lead to 2, 1 or no solution at all. However, when an accurate picture of the triangle or suitable context is available, we can determine which angle is desired.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 455 Try it Now 1. Given α = 80°, a = 120, and b = 121 , find the corresponding and missing side and angles. If there is more than one possible solution, show both. Example 4 Find all possible triangles if one side has length 4 opposite an angle of 50° and a second side has length 10. Using the given information, we can look for the angle opposite the side of length 10. sin(50°) sin(α ) = 10 4 10 sin(50°) sin(α ) = ≈ 1.915 4 Since the range of the sine function is [-1, 1], it is impossible for the sine value to be 1.915. There are no triangles that can be drawn with the provided dimensions. Example 5 Find all possible triangles if one side has length 6 opposite an angle of 50° and a second side has length 4. Using the given information, we can look for the angle opposite the side of length 4. sin(50°) sin(α ) = 6 4 4 sin(50°) Use the inverse to find one solution sin(α ) = ≈ 0.511 6 By symmetry there is a second possible solution α = sin −1 (0.511) ≈ 30.710° α = 180° − 30.710° = 149.290° If we use the angle 30.710° , the third angle would be 180° − 50° − 30.710° = 99.290° . We can then use Law of Sines again to find the third side. sin(50°) sin(99.290°) Solve for c = 6 c c = 7.730 If we used the angle α = 149.290°, the third angle would be 180° – 50° – 149.290° = –19.29°, which is impossible, so the previous triangle is the only possible one. Try it Now 2. Given α = 80°, a = 100, and b = 10 find the missing side and angles. If there is more than one possible solution, show both.
456 Chapter 8 Law of Cosines Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat?
8 mi
Unfortunately, while the Law of Sines lets us address many non-right triangle cases, it does not allow us to address triangles where the one known angle is included between two known sides, which means it is not a corresponding angle for a known side. For this, we need another tool. Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label h, to create two right triangles. We will divide the base b into two pieces, one of which we will temporarily label x. From this picture, we can establish the right triangle relationship x cos(α ) = , or equivalently, x = c cos(α ) c
β
c α
10 mi
a
h x
20°
b
b-x
γ
Using the Pythagorean Theorem, we can establish (b − x )2 + h 2 = a 2 and x 2 + h 2 = c 2 Both of these equations can be solved for h 2 2 and h2 = c2 − x2 h 2 = a 2 − (b − x ) Since the left side of each equation is h 2 , the right sides must be equal 2 Multiply out the right c 2 − x 2 = a 2 − (b − x ) 2 2 2 2 2 Simplify c − x = a − b − 2bx + x
(
2
2
2
)
2
c − x = a − b + 2bx − x c 2 = a 2 − b 2 + 2bx a 2 = c 2 + b 2 − 2bx
2
Isolate a 2 Substitute in c cos(α ) = x from above
a 2 = c 2 + b 2 − 2bc cos(α ) This result is called the Law of Cosines. Depending upon which side we dropped the altitude down from, we could have established this relationship using any of the angles. The important thing to note is that the right side of the equation involves an angle and the sides adjacent to that angle – the left side of the equation involves the side opposite that angle.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 457 Law of Cosines Given a triangle with angles and opposite sides labeled as shown, a 2 = c 2 + b 2 − 2bc cos(α ) b 2 = a 2 + c 2 − 2ac cos( β ) c 2 = a 2 + b 2 − 2ab cos(γ )
γ
a
b α
c
β
Notice that if one of the angles of the triangle is 90 degrees, cos(90°) = 0, so the formula Simplifies to c 2 = a 2 + b 2 − 2ab cos(90°) c2 = a2 + b2 You should recognize this as the Pythagorean Theorem. Indeed, the Law of Cosines is sometimes called the Generalized Pythagorean Theorem, since it extends the Pythagorean Theorem to non-right triangles. Example 6 Returning to our question from earlier, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat?
8 mi
The boat turned 20 degrees, so the obtuse angle of the non-right triangle shown in the picture is the supplemental angle, 180° - 20° = 160°.
20°
10 mi
With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle – the distance from the boat to port. x 2 = 8 2 + 10 2 − 2(8)(10) cos(160°) x 2 = 314.3508 x = 314.3508 = 17.730
Evaluate the cosine and simplify Square root both sides
The boat is 17.73 miles from port. Example 7 Find the unknown side and angles of this triangle. Notice that we don’t have both pieces of any side/angle pair, so the Law of Sines would not work with this triangle.
x θ
φ
10 12
30°
Since we have the angle included between the two known sides, we can turn to Law of Cosines.
458 Chapter 8 Since the left side of any of the Law of Cosines equations involves the side opposite the known angle, the left side in this situation will involve the side x. The other two sides can be used in either order. x 2 = 10 2 + 12 2 − 2(10)(12) cos(30°) x 2 = 10 2 + 12 2 − 2(10)(12)
3 2
x 2 = 244 − 120 3
Evaluate the cosine Simplify Take the square root
x = 244 − 120 3 ≈ 6.013
Now that we know an angle and its corresponding side, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle θ, sin(30°) sin(θ ) = 6.013 10 10 sin(30°) Use the inverse sine sin(θ ) = 6.013 10 sin(30°) θ = sin −1 ≈ 56.256° 6.013 The other possibility for θ would be θ = 180° – 56.256° = 123.744°. In the original picture, θ is an acute angle, so 123.744° doesn’t make sense if we assume the picture is drawn to scale. Proceeding with θ = 56.256°, we can then find the third angle of the triangle: ϕ = 180° − 30° − 56.256° = 93.744° . In addition to solving for the missing side opposite one known angle, the Law of Cosines allows us to find the angles of a triangle when we know all three sides. Example 8 Solve for the angle α in the triangle shown. Using the Law of Cosines, 20 2 = 18 2 + 25 2 − 2(18)(25) cos(α ) 400 = 949 − 900 cos(α ) − 549 = −900 cos(α ) − 549 = cos(α ) − 900 − 549 α = cos −1 ≈ 52.410° − 900
25 Simplify
α 18
20
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 459 Try it Now 3. Given α = 25°, b = 10, and c = 20 find the missing side and angles. Notice that since the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases when using Law of Cosines to find an angle. Example 9 On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is done by a process called triangulation, which works by using the distance from two known points. Suppose there are two cell phone towers within range of you, located 6000 feet apart along a straight highway that runs east to west, and you know you are north of the highway. Based on the signal delay, it can be determined you are 5050 feet from the first tower, and 2420 feet from the second. Determine your position north and east of the first tower, and determine how far you are from the highway. For simplicity, we start by drawing a picture and labeling our given information. Using the Law of Cosines, we can solve for the angle θ. 2420 2 = 6000 2 + 5050 2 − 2(5050)(6000) cos(θ ) 5856400 = 61501500 − 60600000 cos(θ ) − 554646100 = −60600000 cos(θ ) − 554646100 cos(θ ) = = 0.9183 − 60600000 θ = cos −1 (0.9183) = 23.328° Using this angle, we could then use right triangles to find the position of the cell phone relative to the western tower. x 5050 x = 5050 cos(23.328°) ≈ 4637.2 feet y sin( 23.328°) = 5050 y = 5050 sin(23.328°) ≈ 1999.8 feet cos(23.328°) =
5050 ft θ
2420 ft
6000 ft
5050 ft 23.3° x
y
You are 5050 ft from the tower and 23.328° north of east (or, equivalently, 66.672° east of north). Specifically, you are about 4637 feet east and 2000 feet north of the first tower.
460 Chapter 8 Note that if you didn’t know whether you were north or south of the towers, our calculations would have given two possible locations, one north of the highway and one south. To resolve this ambiguity in real world situations, locating a position using triangulation requires a signal from a third tower. Example 10 To measure the height of a hill, a woman measures the angle of elevation to the top of the hill to be 24 degrees. She then moves back 200 feet and measures the angle of elevation to be 22 degrees. Find the height of the hill. As with many problems of this nature, it will be helpful to draw a picture.
22° 200 ft
h
24°
Notice there are three triangles formed here – the right triangle including the height h and the 22 degree angle, the right triangle including the height h and the 24 degree angle, and the (non-right) obtuse triangle including the 200 ft side. Since this is the triangle we have the most information for, we will begin with it. It may seem odd to work with this triangle since it does not include the desired side h, but we don’t have enough information to work with either of the right triangles yet. We can find the obtuse angle of the triangle, since it and the angle of 24 degrees complete a straight line – a 180 degree angle. The obtuse angle must be 180° - 24° = 156°. From this, we can determine that the third angle is 2°. We know one side is 200 feet, and its corresponding angle is 2°, so by introducing a temporary variable x for one of the other sides (as shown below), we can use Law of Sines to solve for this length x. 2° 22°
156° 200 ft
x 200 = sin( 22°) sin( 2°) 200 x = sin( 22°) sin( 2°) x = 2146.77 ft
x 24°
h
Setting up the Law of Sines isolating the x value
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 461 Now that we know x, we can use right triangle properties to solve for h. opposite h h sin( 24°) = = = hypotenuse x 2146.77 h = 2146.77 sin(24°) = 873.17 ft
The hill is 873 feet high. Important Topics of This Section Law of Sines Solving for sides Solving for angles Ambiguous case, 0, 1 or 2 solutions Law of Cosines Solving for sides Solving for angles Generalized Pythagorean Theorem Try it Now Answers
β = 83.2° 1. 1 possible solution γ = 16.8° st
nd
2 solution
β = 96.8° γ = 3.2°
c = 35.2 c = 6.9 If we were given a picture of the triangle it may be possible to eliminate one of these
2. β = 5.65°, γ = 94.35°, c = 101.25 3. β = 21.1°, γ = 133.9°, a = 11.725
462 Chapter 8
Section 8.1 Exercises Solve for the unknown sides and angles of the triangles shown. 10 1.
18 50°
70°
40°
2.
110°
45° 15
25° 6 3.
5
5.
4.
90
6
65°
70°
100
6.
25
7.
75°
120°
30 18
40°
30°
50
8.
Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle for the unknown sides and angles if possible. If there is more than one possible solution, give both. 9. α = 43°, γ = 69°, b = 20
10. α = 35°, γ = 73°, b = 19
11. α = 119°, a = 26, b = 14
12. γ = 113°, b = 10, c = 32
13. β = 50°, a = 105, b = 45
14. β = 67°, a = 49, b = 38
15. α = 43.1°, a = 184.2, b = 242.8
16. α = 36.6°, a = 186.2, b = 242.2
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 463 Solve for the unknown sides and angles of the triangles shown.
20
60°
28
17.
16
30°
10
18. 8 20 10
13 19.
11
5 20.
Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle for the unknown sides and angles if possible. If there is more than one possible solution, give both. 21. γ = 41.2°, a = 2.49, b = 3.13
22. β = 58.7°, a = 10.6, c = 15.7
23. α = 120°, b = 6, c = 7
24. γ = 115°, a = 18, b = 23
25. Find the area of a triangle with sides of length 18, 21, and 32. 26. Find the area of a triangle with sides of length 20, 26, and 37. 27. To find the distance across a small lake, a surveyor has taken the measurements shown. Find the distance across the lake. 800 ft
28. To find the distance between two cities, a satellite calculates the distances and angle shown (not to scale). Find the distance between the cities.
370 km
70°
2.1°
900 ft
350 km
464 Chapter 8 29. To determine how far a boat is from shore, two radar stations 500 feet apart determine the angles out to the boat, as shown. Find the distance of the boat from the station A, and the distance of the boat from shore.
30. The path of a satellite orbiting the earth causes it to pass directly over two tracking stations A and B, which are 69 mi apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 86.2° and 83.9°, respectively. How far is the satellite from station A and how high is the satellite above the ground?
33. A 127 foot tower is located on a hill that is inclined 38° to the horizontal. A guy-wire is to be attached to the top of the tower and anchored at a point 64 feet downhill from the base of the tower. Find the length of wire needed.
B
86.2°
83.9° A
31. A communications tower is located at the top of a steep hill, as shown. The angle of inclination of the hill is 67°. A guy-wire is to be attached to the top of the tower and to the ground, 165 m downhill from the base of the tower. The angle formed by the guy-wire and the hill is 16°. Find the length of the cable required for the guy wire.
32. The roof of a house is at a 20° angle. An 8 foot solar panel is to be mounted on the roof, and should be angled 38° relative to the horizontal for optimal results. How long does the vertical support holding up the back of the panel need to be?
60°
70°
A
B
16° 165m 67°
8 ft 38° 20°
127 ft 64 ft 38°
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 465
34. A 113 foot tower is located on a hill that is inclined 34° to the horizontal. A guy-wire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.
113 ft
98 ft
34°
35. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° and 44°, as shown in the figure. Find the distance of the plane from point A, and the elevation of the plane.
37°
A
36. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in the figure. Find the distance of the plane from point A, and the elevation of the plane.
44°
B
32°
56°
A
B
37. To estimate the height of a building, two students find the angle of elevation from a point (at ground level) down the street from the building to the top of the building is 39°. From a point that is 300 feet closer to the building, the angle of elevation (at ground level) to the top of the building is 50°. If we assume that the street is level, use this information to estimate the height of the building. 38. To estimate the height of a building, two students find the angle of elevation from a point (at ground level) down the street from the building to the top of the building is 35°. From a point that is 300 feet closer to the building, the angle of elevation (at ground level) to the top of the building is 53°. If we assume that the street is level, use this information to estimate the height of the building. 39. A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?
466 Chapter 8 40. Two planes leave the same airport at the same time. One flies at 20 degrees east of north at 500 miles per hour. The second flies at 30 east of south at 600 miles per hour. How far apart are the planes after 2 hours? 41. The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral? 42. The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral? 43. Three circles with radii 6, 7, and 8, all touch as shown. Find the shaded area bounded by the three circles.
44. A rectangle is inscribed in a circle of radius 10 cm as shown. Find the shaded area, inside the circle but outside the rectangle.
55°
Section 8.2 Polar Coordinates 467
Section 8.2 Polar Coordinates The coordinate system we are most familiar with is called the Cartesian coordinate system, a rectangular plane divided into four quadrants by the horizontal and vertical axes. y In earlier chapters, we often found the Cartesian coordinates of a point on a circle at a given angle from the positive horizontal axis. Sometimes that angle, along with the point’s distance from the origin, provides a more useful way of describing the point’s location than conventional Cartesian coordinates. Polar Coordinates Polar coordinates of a point consist of an ordered pair, (r , θ ) , where r is the distance from the point to the origin, and θ is the angle measured in standard position. Notice that if we were to “grid” the plane for polar coordinates, it would look like the graph to the right, with circles at incremental radii, and rays drawn at incremental angles. Example 1
5π Plot the polar point 3, . 6
This point will be a distance of 3 from the origin, at an angle of 5π . Plotting this, 6 Example 2
π Plot the polar point − 2, . 4 Typically we use positive r values, but occasionally we run into cases where r is negative. On a regular number line, we measure positive values to the right and negative values to the left. We will plot this point similarly. To start we rotate to an angle of
π
. 4 Moving this direction, into the first quadrant, would be positive r values. For negative r values, we move the opposite direction, into the third quadrant. Plotting this:
x
468 Chapter 8 5π Note the resulting point is the same as the polar point 2, 4
.
Try it Now 1. Plot the following points given in polar coordinates and label them. π 3π π a. A = 3, b. B = −2, c. C = 4, 3 6 4 Converting Points To convert between polar coordinates and Cartesian coordinates, we recall the relationships we developed back in Chapter 5. Converting Between Polar and Cartesian Coordinates To convert between polar (r , θ ) and Cartesian (x, y) coordinates, we use the relationships x r y sin(θ ) = r y tan(θ ) = x cos(θ ) =
x = r cos(θ )
(x, y) r
y = r sin(θ )
y
θ x
x2 + y2 = r 2
From these relationship and our knowledge of the unit circle, if r = 1 and θ = π polar coordinates would be (r , θ ) = 1, , and the corresponding Cartesian 3 1 3 coordinates ( x, y ) = , . 2 2
π 3
, the
Remembering your unit circle values will come in very handy as you convert between Cartesian and polar coordinates.
Section 8.2 Polar Coordinates 469 Example 3
2π Find the Cartesian coordinates of a point with polar coordinates (r , θ ) = 5, . 3
To find the x and y coordinates of the point, 5 2π 1 x = r cos(θ ) = 5 cos = 5 − = − 2 3 2 3 5 3 2π = y = r sin(θ ) = 5 sin = 5 2 2 3 5 5 3 . The Cartesian coordinates are − , 2 2
Example 4 Find the polar coordinates of the point with Cartesian coordinates (−3,−4) . We begin by finding the distance r using the Pythagorean relationship x 2 + y 2 = r 2 (−3) 2 + (−4) 2 = r 2 9 + 16 = r2 r 2 = 25 r =5 Now that we know the radius, we can find the angle using any of the three trig relationships. Keep in mind that any of the relationships will produce two solutions on the circle, and we need to consider the quadrant to determine which solution to accept. Using the cosine, for example: x −3 cos(θ ) = = r 5 − 3 By symmetry, there is a second possibility at θ = cos −1 ≈ 2.214 5 θ = 2π − 2.214 = 4.069 Since the point (-3, -4) is located in the 3rd quadrant, we can determine that the second angle is the one we need. The polar coordinates of this point are (r , θ ) = (5,4.069) . Try it Now 2. Convert the following. a. Convert polar coordinates (r , θ ) = ( 2, π ) to ( x, y ) . b. Convert Cartesian coordinates ( x, y= ) (0, −4) to (r , θ ) .
470 Chapter 8 Polar Equations Just as a Cartesian equation like y = x 2 describes a relationship between x and y values on a Cartesian grid, a polar equation can be written describing a relationship between r and θ values on the polar grid. Example 5 Sketch a graph of the polar equation r = θ . The equation r = θ describes all the points for which the radius r is equal to the angle. To visualize this relationship, we can create a table of values. θ 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π r 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π We can plot these points on the plane, and then sketch a curve that fits the points. The resulting graph is a spiral. Notice that the resulting graph cannot be the result of a function of the form y = f(x), as it does not pass the vertical line test, even though it resulted from a function giving r in terms of θ. Although it is nice to see polar equations on polar grids, it is more common for polar graphs to be graphed on the Cartesian coordinate system, and so, the remainder of the polar equations will be graphed accordingly. The spiral graph above on a Cartesian grid is shown here. Example 6 Sketch a graph of the polar equation r = 3 . Recall that when a variable does not show up in the equation, it is saying that it does not matter what value that variable has; the output for the equation will remain the same. For example, the Cartesian equation y = 3 describes all the points where y = 3, no matter what the x values are, producing a horizontal line. Likewise, this polar equation is describing all the points at a distance of 3 from the origin, no matter what the angle is, producing the graph of a circle.
Section 8.2 Polar Coordinates 471 The normal settings on graphing calculators and software graph on the Cartesian coordinate system with y being a function of x, where the graphing utility asks for f(x), or simply y =. To graph polar equations, you may need to change the mode of your calculator to Polar. You will know you have been successful in changing the mode if you now have r as a function of θ, where the graphing utility asks for r(θ), or simply r =. Example 7 Sketch a graph of the polar equation r = 4 cos(θ ) , and find an interval on which it completes one cycle. While we could again create a table, plot the corresponding points, and connect the dots, we can also turn to technology to directly graph it. Using technology, we produce the graph shown here, a circle passing through the origin. Since this graph appears to close a loop and repeat itself, we might ask what interval of θ values yields the entire graph. At θ = 0, r = 4 cos(0) = 4 . We would then consider the next θ value when r will be 4, which would mean we are back where we started. Solving, 4 = 4 cos(θ ) cos(θ ) = 1 θ = 0 or θ = π This shows us at 0 radians we are at the point (0, 4), and again at π radians we are at the point (0, 4) having finished one complete revolution. The interval 0 ≤ θ < π yields one complete iteration of the circle. Try it Now 3. Sketch a graph of the polar equation r = 3sin(θ ) , and find an interval on which it completes one cycle. The last few examples have all been circles. Next we will consider two other “named” polar equations, limaçons and roses. Example 8 Sketch a graph of the polar equation r = 4 sin(θ ) + 2 . What interval of θ values corresponds to the inner loop? This type of graph is called a limaçon.
472 Chapter 8 Using technology, we can draw a graph. The inner loop begins and ends at the origin, where r = 0. We can solve for the θ values for which r = 0. 0 = 4 sin(θ ) + 2 − 2 = 4 sin(θ ) 1 sin(θ ) = − 2 11π 7π or θ = θ= 6 6
This tells us that r = 0, or the graph passes through the origin, twice on the interval [0, 2π). 7π 11π The inner loop arises from the interval . This corresponds to where the ≤θ ≤ 6 6 function r = 4 sin(θ ) + 2 takes on negative values. Example 9 Sketch a graph of the polar equation r = cos(3θ ) . What interval of θ values describes one small loop of the graph? This type of graph is called a 3 leaf rose. Again we can use technology to produce a graph. The interval [0, π) yields one cycle of this function. As with the last problem, we can note that there is an interval on which one loop of this graph begins and ends at the origin, where r = 0. Solving for θ, 0 = cos(3θ ) 0 = cos(u )
Substitute u = 3θ
3π 5π or u = 2 2 2 Undo the substitution π 3π or 3θ = 3θ = 2 2 u=
θ=
π
π
6
or u =
or
θ=
π
2
or or
5π 2 5π θ= 6 3θ =
There are 3 solutions on 0 ≤ θ < π which correspond to the 3 times the graph returns to the origin, but the first two solutions we solved for above are enough to conclude that one loop corresponds to the interval
π
6
≤θ <
π
2
.
Section 8.2 Polar Coordinates 473 If we wanted to get an idea of how the computer drew this graph, consider when θ = 0. = r cos(3 = θ ) cos(0) = 1 , so the graph starts at (1,0). As we found above, at θ =
θ=
π 2
, the graph is at the origin. Looking at the equation,
notice that any angle in between
π
π 6
and
π 2
, for example at
π , produces a negative r: r = −1 . cos 3 ⋅ = cos (π ) = 3 3 Notice that with a negative r value and an angle with terminal side in the first quadrant, the corresponding Cartesian point would be in the third quadrant. Since r = cos(3θ ) is negative
θ=
on
π 6
≤θ <
π 2
θ 0
π 6
π
3
π
2
π
and
6
r 1
x 1
y 0
0
0
0
-1
−
0
0
1 2
−
3 2
0
, this interval corresponds to the loop of the graph in the third quadrant.
Try it Now 4. Sketch a graph of the polar equation r = sin(2θ ) . Would you call this function a limaçon or a rose? Converting Equations While many polar equations cannot be expressed nicely in Cartesian form (and vice versa), it can be beneficial to convert between the two forms, when possible. To do this we use the same relationships we used to convert points between coordinate systems. Example 10 Rewrite the Cartesian equation x 2 + y 2 = 6 y as a polar equation. We wish to eliminate x and y from the equation and introduce r and θ. Ideally, we would like to write the equation with r isolated, if possible, which represents r as a function of θ. Remembering x 2 + y 2 = r 2 we substitute x2 + y2 = 6y r2 = 6y y = r sin(θ ) and so we substitute again Subtract 6r sin(θ ) from both sides r 2 = 6r sin(θ ) 2 Factor r − 6r sin(θ ) = 0 Use the zero factor theorem r (r − 6 sin(θ ) ) = 0 Since r = 0 is only a point, we reject that solution. r = 6 sin(θ ) or r = 0 The solution r = 6 sin(θ ) is fairly similar to the one we graphed in Example 7. In fact, this equation describes a circle with bottom at the origin and top at the point (0, 6).
474 Chapter 8 Example 11 Rewrite the Cartesian equation y = 3 x + 2 as a polar equation. y = 3x + 2 r sin(θ ) = 3r cos(θ ) + 2 r sin(θ ) − 3r cos(θ ) = 2 r (sin(θ ) − 3 cos(θ ) ) = 2 2 r= sin(θ ) − 3 cos(θ )
Use y = r sin(θ ) and x = r cos(θ ) Move all terms with r to one side Factor out r Divide
In this case, the polar equation is more unwieldy than the Cartesian equation, but there are still times when this equation might be useful. Example 12 Rewrite the polar equation r =
3 as a Cartesian equation. 1 − 2 cos(θ )
We want to eliminate θ and r and introduce x and y. It is usually easiest to start by clearing the fraction and looking to substitute values that will eliminate θ. 3 Clear the fraction r= 1 − 2 cos(θ ) x Use cos(θ ) = to eliminate θ r (1 − 2 cos(θ ) ) = 3 r x Distribute and simplify r 1 − 2 = 3 r Isolate the r r − 2x = 3 Square both sides r = 3 + 2x 2 2 Use x 2 + y 2 = r 2 r = (3 + 2 x ) x 2 + y 2 = (3 + 2 x )
2
When our entire equation has been changed from r and θ to x and y we can stop unless asked to solve for y or simplify. In this example, if desired, the right side of the equation could be expanded and the equation simplified further. However, the equation cannot be written as a function in Cartesian form. Try it Now 5. a. Rewrite the Cartesian equation in polar form: y = ± 3 − x2 b. Rewrite the polar equation in Cartesian form: r = 2sin(θ )
Section 8.2 Polar Coordinates 475 Example 13 Rewrite the polar equation r = sin(2θ ) in Cartesian form. r = sin(2θ )
Use the double angle identity for sine y x Use cos(θ ) = and sin(θ ) = r r
r = 2 sin(θ ) cos(θ )
x y r = 2⋅ ⋅ r r 2 xy r= 2 r
r 3 = 2 xy
(x
2
+ y2
Simplify Multiply by r2 Since x 2 + y 2 = r 2 , r = x 2 + y 2
) = 2 xy 3
This equation could also be written as
(x
2
+ y2
)
3/ 2
= 2 xy
x 2 + y 2 = (2xy )
or
2/3
Important Topics of This Section Cartesian coordinate system Polar coordinate system Plotting points in polar coordinates Converting coordinates between systems Polar equations: Spirals, circles, limaçons and roses Converting equations between systems Try it Now Answers C A
B 1. 2. a. (r , θ ) = ( 2, π ) converts to ( x, y ) = (−2, 0) b. ( x, y= )
to (r , θ ) = ( 0, −4 ) converts
3π 4, 2
π or −4, 2
476 Chapter 8
3.
4.
It completes one cycle on the interval 0 ≤ θ < π .
This is a 4-leaf rose.
5. a. y = ± 3 − x 2 becomes r = 3 b. r = 2sin(θ ) becomes x 2 + y 2 = 2y
Section 8.2 Polar Coordinates 477
Section 8.2 Exercises Convert the given polar coordinates to Cartesian coordinates. 7π 7π 3π 1. 7, 2. 6, 3. 4, 6 4 4
4π 4. 9, 3
π 5. 6, − 4
π 6. 12, − 3
π 7. 3, 2
8. ( 5, π )
π 9. −3, 6
2π 10. −2, 3
11. (3, 2)
12. (7,1)
Convert the given Cartesian coordinates to polar coordinates. 14. (8, 8) 15. (−4, 6) 13. (4, 2) 17. (3, 5) −
18. (6, 5) −
19. ( −10, 13 − )
16. (−5,1 ) 20. (−4, 7) −
Convert the given Cartesian equation to a polar equation. 22. y = 4 23. y = 4 x 2 21. x = 3
24. y = 2 x 4
25. x 2 + y 2 = 4y
28. x 2 − y 2 = 3y
26. x 2 + y 2 = 3x
27. x 2 − y 2 = x
Convert the given polar equation to a Cartesian equation. 30. r = 4 cos (θ ) 29. r = 3sin (θ ) 31. r =
4 sin (θ ) + 7 cos (θ )
33. r = 2sec (θ ) 35. r =
r cos (θ ) + 2
32. r =
6 cos (θ ) + 3sin (θ )
34. r = 3csc (θ ) 36. r 2 = 4sec (θ ) csc (θ )
478 Chapter 8 Match each equation with one of the graphs shown. 37. r= 2 + 2 cos (θ ) 38. r= 2 + 2sin (θ )
39. r= 4 + 3cos (θ )
40. r= 3 + 4 cos (θ )
42. r = 2sin (θ )
41. r = 5
A
B
C
D
E
F
Match each equation with one of the graphs shown.
θ 45. r = cos 2
43. r = log (θ )
44. r = θ cos (θ )
46. r = sin (θ ) cos 2 (θ )
47. r = 1 + 2sin ( 3θ ) 48. r = 1 + sin ( 2θ )
A
B
C
D
E
F
Section 8.2 Polar Coordinates 479 Sketch a graph of the polar equation. 49. r = 3cos (θ ) 50. r = 4sin (θ )
51. r = 3sin ( 2θ )
52. r = 4sin ( 4θ )
53. r = 5sin ( 3θ )
54. r = 4sin ( 5θ )
55. r = 3cos ( 2θ )
56. r = 4 cos ( 4θ )
57. r= 2 + 2 cos (θ )
58. r= 3 + 3sin (θ )
59. r = 1 + 3sin (θ )
60. r= 2 + 4 cos (θ )
61. r = 2θ
62. r =
1
θ 1
63. r= 3 + sec (θ ) , a conchoid
64. r =
65. r = 2sin (θ ) tan (θ ) , a cissoid
66.= r 2 1 − sin 2 (θ ) , a hippopede
1
θ
, a lituus 1
This curve was the inspiration for the artwork featured on the cover of this book.
480 Chapter 8
Section 8.3 Polar Form of Complex Numbers From previous classes, you may have encountered “imaginary numbers” – the square roots of negative numbers – and, more generally, complex numbers which are the sum of a real number and an imaginary number. While these are useful for expressing the solutions to quadratic equations, they have much richer applications in electrical engineering, signal analysis, and other fields. Most of these more advanced applications rely on properties that arise from looking at complex numbers from the perspective of polar coordinates. We will begin with a review of the definition of complex numbers. Imaginary Number i The most basic complex number is i, defined to be i = − 1 , commonly called an imaginary number. Any real multiple of i is also an imaginary number. Example 1 Simplify − 9 . We can separate − 9 as square root of -1 as i. − 9 = 9 − 1 = 3i
9 − 1 . We can take the square root of 9, and write the
A complex number is the sum of a real number and an imaginary number. Complex Number A complex number is a number z = a + bi , where a and b are real numbers a is the real part of the complex number b is the imaginary part of the complex number i = −1 Plotting a complex number We can plot real numbers on a number line. For example, if we wanted to show the number 3, we plot a point:
Section 8.3 Polar Form of Complex Numbers 481 To plot a complex number like 3 − 4i , we need more than just a number line since there are two components to the number. To plot this number, we need two number lines, crossed to form a complex plane. Complex Plane In the complex plane, the horizontal axis is the real axis and the vertical axis is the imaginary axis.
Example 2 Plot the number 3 − 4i on the complex plane.
imaginary real
imaginary
The real part of this number is 3, and the imaginary part is -4. To plot this, we draw a point 3 units to the right of the origin in the horizontal direction and 4 units down in the vertical direction.
real
Because this is analogous to the Cartesian coordinate system for plotting points, we can think about plotting our complex number z = a + bi as if we were plotting the point (a, b) in Cartesian coordinates. Sometimes people write complex numbers as z= x + yi to highlight this relation. Arithmetic on Complex Numbers Before we dive into the more complicated uses of complex numbers, let’s make sure we remember the basic arithmetic involved. To add or subtract complex numbers, we simply add the like terms, combining the real parts and combining the imaginary parts. Example 3 Add 3 − 4i and 2 + 5i . Adding (3 − 4i ) + (2 + 5i ) , we add the real parts and the imaginary parts 3 + 2 − 4i + 5i 5+i Try it Now 1. Subtract 2 + 5i from 3 − 4i . We can also multiply and divide complex numbers.
482 Chapter 8 Example 4 Multiply: 4(2 + 5i ) . To multiply the complex number by a real number, we simply distribute as we would when multiplying polynomials. 4(2 + 5i ) = 4 ⋅ 2 + 4 ⋅ 5i = 8 + 20i
Example 5 (2 + 5i ) Divide . (4 − i ) To divide two complex numbers, we have to devise a way to write this as a complex number with a real part and an imaginary part. We start this process by eliminating the complex number in the denominator. To do this, we multiply the numerator and denominator by a special complex number so that the result in the denominator is a real number. The number we need to multiply by is called the complex conjugate, in which the sign of the imaginary part is changed. Here, 4+i is the complex conjugate of 4–i. Of course, obeying our algebraic rules, we must multiply by 4+i on both the top and bottom. (2 + 5i ) (4 + i ) ⋅ (4 − i ) (4 + i ) To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL – “first outer inner last”). In the numerator: (2 + 5i )(4 + i )
Expand
=8 + 20i + 2i + 5i 2 =8 + 20i + 2i + 5(−1) = 3 + 22i
Since i = − 1 , i 2 = −1 Simplify
Following the same process to multiply the denominator Expand (4 − i )(4 + i ) (16 − 4i + 4i − i 2 ) (16 − (−1)) =17 Combining this we get
Since i = − 1 , i 2 = −1
3 + 22i 3 22i = + 17 17 17
Section 8.3 Polar Form of Complex Numbers 483 Try it Now 2. Multiply 3 − 4i and 2 + 3i . With the interpretation of complex numbers as points in a plane, which can be related to the Cartesian coordinate system, you might be starting to guess our next step – to refer to this point not by its horizontal and vertical components, but using its polar location, given by the distance from the origin and an angle. Polar Form of Complex Numbers Remember, because the complex plane is analogous to the Cartesian plane that we can think of a complex number z= x + yi as analogous to the Cartesian point (x, y) and recall how we converted from (x, y) to polar (r, θ) coordinates in the last section. Bringing in all of our old rules we remember the following: x cos(θ ) = r y sin(θ ) = r y tan(θ ) = x
imaginary x = r cos(θ ) y = r sin(θ ) 2
2
x +y =r
x + yi r
2
θ x
y
real
With this in mind, we can write z =x + yi =r cos(θ ) + ir sin(θ ) . Example 6 Express the complex number 4i using polar coordinates. On the complex plane, the number 4i is a distance of 4 from π π π the origin at an angle of , so 4i = 4 cos + i 4 sin 2 2 2
4i 4
π 2
Note that the real part of this complex number is 0. In the 18th century, Leonhard Euler demonstrated a relationship between exponential and trigonometric functions that allows the use of complex numbers to greatly simplify some trigonometric calculations. While the proof is beyond the scope of this class, you will likely see it in a later calculus class.
484 Chapter 8 Polar Form of a Complex Number and Euler’s Formula The polar form of a complex number is z = re iθ , where Euler’s Formula holds: re iθ = r cos(θ ) + ir sin(θ ) Similar to plotting a point in the polar coordinate system we need r and θ to find the polar form of a complex number. Example 7 Find the polar form of the complex number -7. Treating this is a complex number, we can consider the unsimplified version -7+0i. Plotted in the complex plane, the number -7 is on the negative horizontal axis, a distance of 7 from the origin at an angle of π from the positive horizontal axis. The polar form of the number -7 is 7e iπ . Plugging r = 7 and θ = π back into Euler’s formula, we have: 7e iπ = 7 cos(π ) + 7i sin(π ) = −7 + 0i = −7 as desired. Example 8 Find the polar form of − 4 + 4i . On the complex plane, this complex number would correspond to the point (-4, 4) on a Cartesian plane. We can find the distance r and angle θ as we did in the last section. r 2 = x2 + y2 r 2 = (−4) 2 + 4 2 r = 32 = 4 2 To find θ, we can use cos(θ ) = −4
x r
-4+4i
2 2 4 2 This is one of known cosine values, and since the point is in 3π the second quadrant, we can conclude that θ = . 4 cos(θ ) =
=−
The polar form of this complex number is 4 2e
3π i 4
.
4 2
3π 4
Section 8.3 Polar Form of Complex Numbers 485 Note we could have used tan(θ ) = check the quadrant.
y instead to find the angle, so long as we remember to x
Try it Now 3. Write 3 + i in polar form. Example 9
π
i
Write 3e 6 in complex a + bi form. π
i π π 3e 6 = 3 cos + i3 sin 6 6 3 1 = 3⋅ + i3 ⋅ 2 2 3 3 3 = +i 2 2
Evaluate the trig functions Simplify
The polar form of a complex number provides a powerful way to compute powers and roots of complex numbers by using exponent rules you learned in algebra. To compute a power of a complex number, we: 1) Convert to polar form 2) Raise to the power, using exponent rules to simplify 3) Convert back to a + bi form, if needed Example 10 6 Evaluate (− 4 + 4i ) . While we could multiply this number by itself five times, that would be very tedious. To compute this more efficiently, we can utilize the polar form of the complex number. In an earlier example, we found that − 4 + 4i = 4 2e
(− 4 + 4i )6 3π i = 4 2e 4
( )
. Using this,
Write the complex number in polar form 6
34π i = 4 2 e 6
3π i 4
Utilize the exponent rule (ab) m = a m b m 6
On the second factor, use the rule (a m ) n = a mn
486 Chapter 8
( )
6
= 4 2 e = 32768e
3π i ⋅6 4
Simplify
9π i 2
At this point, we have found the power as a complex number in polar form. If we want the answer in standard a + bi form, we can utilize Euler’s formula. 32768e
9π i 2
9π 9π = 32768 cos + i32768 sin 2 2
9π π is coterminal with , we can use our special angle knowledge to evaluate 2 2 the sine and cosine. 9π 9π 32768 cos + i32768 sin = 32768(0) + i32768(1) = 32768i 2 2
Since
We have found that (− 4 + 4i ) = 32768i . 6
The result of the process can be summarized by DeMoivre’s Theorem. DeMoivre’s Theorem If z r ( cos (θ ) + i sin (θ ) ) , then for any integer n, z n r n ( cos ( nθ ) + i sin ( nθ ) ) = = We omit the proof, but note we can easily verify it holds in one case using Example 10: 6 9π 3π 9π 3π (−4 + 4i ) 6 = 4 2 cos 6 ⋅ + i sin 6 ⋅ = 32768 cos + i sin = 32768i 4 4 2 2
( )
Example 11 Evaluate 9i . To evaluate the square root of a complex number, we can first note that the square root 1 is the same as having an exponent of : 9i = (9i )1 / 2 . 2 To evaluate the power, we first write the complex number in polar form. Since 9i has no real part, we know that this value would be plotted along the vertical axis, a distance of 9 from the origin at an angle of
π 2
π
i
. This gives the polar form: 9i = 9e 2 .
Section 8.3 Polar Form of Complex Numbers 487 9i = (9i )1 / 2
πi = 9e 2 =9
1/ 2
=9
1/ 2
π
= 3e 4
Use the polar form
1/ 2
Use exponent rules to simplify
π2 i e
1/ 2
π 1
e2
i⋅
Simplify
2
i
π π = 3 cos + i3 sin 4 4 2 2 =3 + i3 2 2
Rewrite using Euler’s formula if desired Evaluate the sine and cosine
Using the polar form, we were able to find a square root of a complex number. 3 2 3 2 9i = + i 2 2 Alternatively, using DeMoivre’s Theorem we can write 1/ 2
π2 i π π 9e = 3 cos + i sin and simplify 4 4 Try it Now 4. Write
(
3 +i
)
6
in polar form.
You may remember that equations like x 2 = 4 have two solutions, 2 and -2 in this case, though the square root 4 only gives one of those solutions. Likewise, the square root we found in Example 11 is only one of two complex numbers whose square is 9i. Similarly, the equation z 3 = 8 would have three solutions where only one is given by the cube root. In this case, however, only one of those solutions, z = 2, is a real value. To find the others, we can use the fact that complex numbers have multiple representations in polar form. Example 12 Find all complex solutions to z 3 = 8 .
488 Chapter 8 Since we are trying to solve z 3 = 8 , we can solve for x as z = 81/ 3 . Certainly one of these solutions is the basic cube root, giving z = 2. To find others, we can turn to the polar representation of 8. Since 8 is a real number, is would sit in the complex plane on the horizontal axis at an angle of 0, giving the polar form 8e 0i . Taking the 1/3 power of this gives the real solution:
(8e )
0i 1 / 3
( )
= 81 / 3 e 0i
1/ 3
= 2e 0 = 2 cos(0) + i 2 sin(0) = 2
However, since the angle 2π is coterminal with the angle of 0, we could also represent the number 8 as 8e 2πi . Taking the 1/3 power of this gives a first complex solution: 2π 2πi 1 / 3 1/ 3 2πi 1 / 3 (8e ) = 8 (e ) = 2e 3 i = 2 cos 23π + i 2 sin 23π = 2 − 12 + i 2 23 = −1 + 3i To find the third root, we use the angle of 4π, which is also coterminal with an angle of 0. 4π 4πi 1 / 3 1/ 3 4πi 1 / 3 (8e ) = 8 (e ) = 2e 3 i = 2 cos 43π + i 2 sin 43π = 2 − 12 + i 2 − 23 = −1 − 3i 3 Altogether, we found all three complex solutions to z = 8 , − 1 + 3i = z 2, − 1 + 3i, − 1 − 3i 2
Graphed, these three numbers would be equally spaced on a circle about the origin at a radius of 2. − 1 − 3i Important Topics of This Section Complex numbers Imaginary numbers Plotting points in the complex coordinate system Basic operations with complex numbers Euler’s Formula DeMoivre’s Theorem Finding complex solutions to equations Try it Now Answers 1. (3 − 4i ) − (2 + 5i ) =1 − 9i 2. (3 − 4i )(2 + 3i ) =18 + i 3. 3 + i in polar form is 2e 4. −64
iπ
6
Section 8.3 Polar Form of Complex Numbers 489
Section 8.3 Exercises
Simplify each expression to a single complex number. 1. −9 2. −16 4.
5.
−3 −75
2 + −12 2
3. 6.
−6 −24 4 + −20 2
Simplify each expression to a single complex number. 7. ( 3 + 2i ) + (5 − 3i ) 8. ( −2 − 4i ) + (1 + 6i ) 9. ( −5 + 3i ) − (6 − i )
10. ( 2 − 3i ) − (3 + 2i )
11. ( 2 + 3i ) (4i )
12. ( 5 − 2i ) (3i )
13. ( 6 − 2i ) (5)
14. ( −2 + 4i )( 8 )
15. ( 2 + 3i ) (4 − i )
16. ( −1 + 2i ) (−2 + 3i )
17. ( 4 − 2i ) (4 + 2i )
18. ( 3 + 4i )( 3 − 4i )
19.
3 + 4i 2
20.
6 − 2i 3
21.
−5 + 3i 2i
22.
6 + 4i i
23.
2 − 3i 4 + 3i
24.
3 + 4i 2−i
25. i 6
26. i11
27. i17
28. i 24
Rewrite each complex number from polar form into a + bi form. 29. 3e 33. 3e
2i 5π i 4
30. 4e 34. 5e
4i
31. 6e
π
6
i
π
32. 8e 3
i
7π i 4
Rewrite each complex number into polar reiθ form. 35. 6 36. −8 37. −4i
38. 6i
39. 2 + 2i
40. 4 + 4i
41. −3 + 3i
42. −4 − 4i
43. 5 + 3i
44. 4 + 7i
45. −3 + i
46. −2 + 3i
47. −1 − 4i
48. −3 − 6i
49. 5 − i
50. 1 − 3i
490 Chapter 8
Compute each of the following, leaving the result in polar reiθ form.
π i π4 i 51. 3e 6 2e 4π
54.
24e 3 π
6e 2
10
i
πi 55. 2e 4
i
2π
2π i 5π i 52. 2e 3 4e 3
i
57. 16e 3
3π
58. 9e 2
3π
53.
6e 4 π
3e 6
i
i
πi 56. 3e 6
4
i
Compute each of the following, simplifying the result into a + bi form. 8 6 60. ( 4 + 4i ) 61. −3 + 3i 59. ( 2 + 2i ) 62.
−4 − 4i
63.
3
5 + 3i
64.
Solve each of the following equations for all complex solutions. 65. z 5 = 2 66. z 7 = 3 67. z 6 = 1
4
4 + 7i
68. z 8 = 1
Section 8.4 Vectors 491
Section 8.4 Vectors A woman leaves home, walks 3 miles north, then 2 miles southeast. How far is she from home, and in which direction would she need to walk to return home? How far has she walked by the time she gets home? This question may seem familiar – indeed we did a similar problem with a boat in the first section of this chapter. In that section, we solved the problem using triangles. In this section, we will investigate another way to approach the problem using vectors, a geometric entity that indicates both a distance and a direction. We will begin our investigation using a purely geometric view of vectors. A Geometric View of Vectors Vector A vector is an object that has both a length and a direction. Geometrically, a vector can be represented by an arrow that has a fixed length and indicates a direction. If, starting at the point A, a vector, which means “carrier” in
Latin, moves toward point B, we write AB to represent the vector.
A vector may also be indicated using a single letter in boldface type, like u, or by capping the letter representing the vector with an arrow, like u . Example 1 Find a vector that represents the movement from the point P(-1, 2) to the point Q(3,3) By drawing an arrow from the first point to the second, we can construct a vector PQ .
Q P
Using this geometric representation of vectors, we can visualize the addition and scaling of vectors. To add vectors, we envision a sum of two movements. To find u + v , we first draw the vector u , then from the end of u we drawn the vector v . This corresponds to the notion that first we move along the first vector, and then from that end position we move along the second vector. The sum u + v is the new vector that travels directly from the beginning of u to the end of v in a straight path.
492 Chapter 8 Adding Vectors Geometrically To add vectors geometrically, draw v starting from the end of u . The sum u + v is the vector from the beginning of u to the end of v .
v u +v
u
Example 2 Given the two vectors shown below, draw u + v
u
v
We draw v starting from the end of u , then draw in the sum u + v from the beginning of u to the end of v .
v u
u +v
Notice that path of the walking woman from the beginning of the section could be visualized as the sum of two vectors. The resulting sum vector would indicate her end position relative to home. Try it Now 1. Draw a vector, v , that travels from the origin to the point (3, 5). Note that although vectors can exist anywhere in the plane, if we put the starting point at the origin it is easy to understand its size and direction relative to other vectors. To scale vectors by a constant, such as 3u , we can imagine adding u + u + u . The result will be a vector three times as long in the same direction as the original vector. If we were to scale a vector by a negative number, such as − u , we can envision this as the opposite of u ; the vector so that u + (−u ) returns us to the starting point. This vector − u would point in the opposite direction as u but have the same length.
Another way to think about scaling a vector is to maintain its direction and multiply its length by a constant, so that 3u would point in the same direction but will be 3 times as long.
Section 8.4 Vectors 493 Scaling a Vector Geometrically To geometrically scale a vector by a constant, scale the length of the vector by the constant. Scaling a vector by a negative constant will reverse the direction of the vector. Example 3 Given the vector shown, draw 3u , − u , and − 2u .
u
The vector 3u will be three times as long. The vector − u will have the same length but point in the opposite direction. The vector − 2u will point in the opposite direction and be twice as long.
3u
−u
−2u
By combining scaling and addition, we can find the difference between vectors geometrically as well, since u − v = u + (−v ) . Example 4 Given the vectors shown, draw u − v
u
v
From the end of u we draw − v , then draw in the result.
Notice that the sum and difference of two vectors are the two diagonals of a parallelogram with the vectors u and v as edges.
Try it Now 2. Using vector v from Try it Now #1, draw − 2v .
−v u −v
u
v
u −v u +v u
v
u
494 Chapter 8 Component Form of Vectors While the geometric interpretation of vectors gives us an intuitive understanding of vectors, it does not provide us a convenient way to do calculations. For that, we need a handy way to represent vectors. Since a vector involves a length and direction, it would be logical to want to represent a vector using a length and an angle θ, usually measured from standard position. u θ
Magnitude and Direction of a Vector A vector u can be described by its magnitude, or length, u , and an angle θ. A vector with length 1 is called unit vector. While this is very reasonable, and a common way to describe vectors, it is often more convenient for calculations to represent a vector by horizontal and vertical components. Component Form of a Vector The component form of a vector represents the vector using two components. u = x, y indicates the vector represents a displacement of x units horizontally and y units vertically. u θ x
y
Notice how we can see the magnitude of the vector as the length of the hypotenuse of a right triangle, or in polar form as the radius, r. Alternate Notation for Vector Components Sometimes you may see vectors written as the combination of unit vectors i and j , where i points the right and j points up. In other words, i = 1,0 and j = 0,1 . In this notation, the vector u = 3,−4 would be written as u = 3i − 4 j since both indicate a displacement of 3 units to the right, and 4 units down.
While it can be convenient to think of the vector u = x, y as an arrow from the origin to the point (x, y), be sure to remember that most vectors can be situated anywhere in the plane, and simply indicate a displacement (change in position) rather than a position.
Section 8.4 Vectors 495 It is common to need to convert from a magnitude and angle to the component form of the vector and vice versa. Happily, this process is identical to converting from polar coordinates to Cartesian coordinates, or from the polar form of complex numbers to the a+bi form. Example 5 Find the component form of a vector with length 7 at an angle of 135 degrees. Using the conversion formulas x = r cos(θ ) and y = r sin(θ ) , we can find the components 7 2 x = 7 cos(135°) = − 2 7 2 y = 7 sin(135°) = 2 This vector can be written in component form as −
7 2 7 2 . , 2 2
Example 6 Find the magnitude and angle θ representing the vector u = 3,−2 . First we can find the magnitude by remembering the relationship between x, y and r: r 2 = 3 2 + (−2) 2 = 13 r = 13 Second we can find the angle. Using the tangent, −2 tan(θ ) = 3 2 θ = tan −1 − ≈ −33.69° , or written as a coterminal positive angle, 326.31°, because 3 we know our point lies in the 4th quadrant.
Try it Now 3. Using vector v from Try it Now #1, the vector that travels from the origin to the point (3, 5), find the components, magnitude and angle θ that represent this vector.
496 Chapter 8 In addition to representing distance movements, vectors are commonly used in physics and engineering to represent any quantity that has both direction and magnitude, including velocities and forces. Example 7 An object is launched with initial velocity 200 meters per second at an angle of 30 degrees. Find the initial horizontal and vertical velocities. By viewing the initial velocity as a vector, we can resolve the vector into horizontal and vertical components. 3 x = 200 cos(30°) = 200 ⋅ ≈ 173.205 m/sec 200 m/s 2 100 m/s 30° 1 173 m/s y = 200 sin(30°) = 200 ⋅ = 100 m/sec 2 This tells us that, absent wind resistance, the object will travel horizontally at about 173 meters each second. Gravity will cause the vertical velocity to change over time – we’ll leave a discussion of that to physics or calculus classes. Adding and Scaling Vectors in Component Form To add vectors in component form, we can simply add the corresponding components. To scale a vector by a constant, we scale each component by that constant. Combining Vectors in Component Form To add, subtract, or scale vectors in component form If u = u1 ,u 2 , v = v1 , v 2 , and c is any constant, then u + v = u1 + v1 , u 2 + v 2 u − v = u1 − v1 , u 2 − v 2 cu = cu1 , cu 2 Example 8 Given u = 3,−2 and v = − 1,4 , find a new vector w = 3u − 2v Using the vectors given, w = 3u − 2v = 3 3,−2 − 2 − 1,4 Scale each vector
= 9,−6 − − 2,8 = 11,−14
Subtract corresponding components
Section 8.4 Vectors 497 By representing vectors in component form, we can find the resulting displacement vector after a multitude of movements without needing to draw a lot of complicated nonright triangles. For a simple example, we revisit the problem from the opening of the section. The general procedure we will follow is: 1) Convert vectors to component form 2) Add the components of the vectors 3) Convert back to length and direction if needed to suit the context of the question Example 9 A woman leaves home, walks 3 miles north, then 2 miles southeast. How far is she from home, and what direction would she need to walk to return home? How far has she walked by the time she gets home? Let’s begin by understanding the question in a little more depth. When we use vectors to describe a traveling direction, we often position things so north points in the upward direction, east points to the right, and so on, as pictured here. Consequently, travelling NW, SW, NE or SE, means we are travelling through the quadrant bordered by the given directions at a 45 degree angle.
NW
N
NE E
W SW
SE
S
With this in mind, we begin by converting each vector to components. A walk 3 miles north would, in components, be 0,3 . A walk of 2 miles southeast would be at an angle of 45° South of East, or measuring from standard position the angle would be 315°. Converting to components, we choose to use the standard position angle so that we do not have to worry about whether the signs are negative or positive; they will work out automatically. 2 cos(315°), 2 sin(315°) = 2 ⋅
2 − 2 , 2⋅ ≈ 1.414, − 1.414 2 2
Adding these vectors gives the sum of the movements in component form 0,3 + 1.414,−1.414 = 1.414,1.586
2 3
To find how far she is from home and the direction she would need to walk to return home, we could find the magnitude and angle of this vector. Length = 1.414 2 + 1.586 2 = 2.125
498 Chapter 8 To find the angle, we can use the tangent 1.586 tan(θ ) = 1.414 1.586 θ = tan −1 = 48.273° north of east 1.414 Of course, this is the angle from her starting point to her ending point. To return home, she would need to head the opposite direction, which we could either describe as 180°+48.273° = 228.273° measured in standard position, or as 48.273° south of west (or 41.727° west of south). She has walked a total distance of 3 + 2 + 2.125 = 7.125 miles. Keep in mind that total distance travelled is not the same as the length of the resulting displacement vector or the “return” vector. Try it Now 4. In a scavenger hunt, directions are given to find a buried treasure. From a starting point at a flag pole you must walk 30 feet east, turn 30 degrees to the north and travel 50 feet, and then turn due south and travel 75 feet. Sketch a picture of these vectors, find their components, and calculate how far and in what direction you must travel to go directly to the treasure from the flag pole without following the map. While using vectors is not much faster than using law of cosines with only two movements, when combining three or more movements, forces, or other vector quantities, using vectors quickly becomes much more efficient than trying to use triangles. Example 10 Three forces are acting on an object as shown below, each measured in Newtons (N). What force must be exerted to keep the object in equilibrium (where the sum of the forces is zero)? 6N
7N
30° 300° 4N
We start by resolving each vector into components.
Section 8.4 Vectors 499 The first vector with magnitude 6 Newtons at an angle of 30 degrees will have components 6 cos(30°), 6 sin(30°) = 6 ⋅
3 1 ,6⋅ = 3 3, 3 2 2
The second vector is only in the horizontal direction, so can be written as − 7, 0 . The third vector with magnitude 4 Newtons at an angle of 300 degrees will have components − 3 1 = 2, − 2 3 4 cos(300°), 4 sin(300°) = 4 ⋅ , 4 ⋅ 2 2
To keep the object in equilibrium, we need to find a force vector x, y so the sum of the four vectors is the zero vector, 0,0 . 3 3, 3 + −7, 0 + 2, − 2 3 + x, y = 0, 0
Add component-wise
3 3 − 7 + 2, 3 + 0 − 2 3 + x, y =0, 0
Simplify
3 3 − 5, 3 − 2 3 + x, y = 0, 0
Solve
x, y =
0, 0 − 3 3 − 5, 3 − 2 3
x, y = −3 3 + 5, − 3 + 2 3 ≈ −0.196, 0.464 This vector gives in components the force that would need to be applied to keep the object in equilibrium. If desired, we could find the magnitude of this force and direction it would need to be applied in. Magnitude = (−0.196) 2 + 0.464 2 = 0.504 N Angle: 0.464 tan(θ ) = − 0.196 0.464 θ = tan −1 = −67.089° . − 0.196 This is in the wrong quadrant, so we adjust by finding the next angle with the same tangent value by adding a full period of tangent: θ = −67.089° + 180° = 112.911° To keep the object in equilibrium, a force of 0.504 Newtons would need to be applied at an angle of 112.911°.
500 Chapter 8 Important Topics of This Section Vectors, magnitude (length) & direction Addition of vectors Scaling of vectors Components of vectors Vectors as velocity Vectors as forces Adding & Scaling vectors in component form Total distance travelled vs. total displacement Try it Now Answers
v
−2v 1
2.
3. v = 3, 5
magnitude = 34
4.
50 ft
30 ft
5 3
θ = tan −1 = 59.04°
75 ft
v1 = 30, 0 v 2 = 50 cos(30°), 50 sin(30°) v3 = 0, − 75 v f = 30 + 50 cos(30°), 50 sin(30°) − 75 = 73.301, − 50 Magnitude = 88.73 feet at an angle of 34.3° south of east.
Section 8.4 Vectors 501
Section 8.4 Exercises Write the vector shown in component form.
1.
2.
Given the vectors shown, sketch u + v , u − v , and 2u .
4.
3.
Write each vector below as a combination of the vectors u and v from question #3.
5.
6.
From the given magnitude and direction in standard position, write the vector in component form. 7. Magnitude: 6, Direction: 45° 8. Magnitude: 10, Direction: 120° 9. Magnitude: 8, Direction: 220° 10. Magnitude: 7, Direction: 305° Find the magnitude and direction of the vector. 11. 0, 4 12. − 3, 0 13. 6, 5 15. − 2,1
16. − 10,13
19. − 4, − 6
20. − 1, 9
17. 2, − 5
14. 3, 7 18. 8, − 4
Using the vectors given, compute u + v , u − v , and 2u − 3v . 21. u = 2, − 3 , v = 1, 5 22. u = − 3, 4 , v = − 2,1
502 Chapter 8 23. A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home? 24. A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina? 25. A person starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far have they walked? If they walked straight home, how far would they have to walk? 26. A person starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far have they walked? If they walked straight home, how far would they have to walk? 27. Three forces act on an object: F1 = − 8, − 5 , F2 = 0,1 , F3 = 4, − 7 . Find the net force acting on the object. 28. Three forces act on an object: F1 = 2, 5 , F2 = 8, 3 , F3 = 0, − 7 . Find the net force acting on the object. 29. A person starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If they walked straight home, how far would they have to walk, and in what direction? 30. A person starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If they walked straight home, how far would they have to walk, and in what direction? 31. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 32. An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? 33. An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, the pilot will need to fly the plane how many degrees west of north?
Section 8.4 Vectors 503
34. An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, the pilot will need to fly the plane how many degrees west of north? 35. As part of a video game, the point (5, 7) is rotated counterclockwise about the origin through an angle of 35 degrees. Find the new coordinates of this point. 36. As part of a video game, the point (7, 3) is rotated counterclockwise about the origin through an angle of 40 degrees. Find the new coordinates of this point. 37. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is travelling 25 mph down the road. If one child doesn't catch the ball and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? 38. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is travelling 45 mph down the road. If one child doesn't catch the ball and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?
504 Chapter 8
Section 8.5 Parametric Equations Many shapes, even ones as simple as circles, cannot be represented as an equation where y is a function of x. Consider, for example, the path a moon follows as it orbits around a planet, which simultaneously rotates around a sun. In some cases, polar equations provide a way to represent such a path. In others, we need a more versatile approach that allows us to represent both the x and y coordinates in terms of a third variable, or parameter. Parametric Equations A system of parametric equations is a pair of functions x(t) and y(t) in which the x and y coordinates are the output, represented in terms of a third input parameter, t. Example 1 Moving at a constant speed, an object moves at a steady rate along a straight path from coordinates (-5, 3) to the coordinates (3, -1) in 4 seconds, where the coordinates are measured in meters. Find parametric equations for the position of the object. The x coordinate of the object starts at -5 meters, and goes to +3 meters, this means the x direction has changed by 8 meters in 4 seconds, giving us a rate of 2 meters per second. We can now write the x coordinate as a linear function with respect to time, t, x(t ) = −5 + 2t . Similarly, the y value starts at 3 and goes to -1, giving a change in y value of 4 meters, meaning the y values have decreased by 4 meters in 4 seconds, for a rate of -1 meter per second, t x y giving equation y (t ) = 3 − t . Together, these are the 0 -5 3 parametric equations for the position of the object: 1 -3 2 x(t ) = −5 + 2t 2 -1 1 3 1 0 y (t ) = 3 − t 4 3 -1 Using these equations, we can build a table of t, x, and y values. Because of the context, we limited ourselves to non-negative t values for this example, but in general you can use any values. From this table, we could create three possible graphs: a graph of x vs. t, which would show the horizontal position over time, a graph of y vs. t, which would show the vertical position over time, or a graph of y vs. x, showing the position of the object in the plane.
Section 8.5 Parametric Equations 505 Position of x as a function of time
Position of y as a function of time y
x
t
t
Position of y relative to x y
x
Notice that the parameter t does not explicitly show up in this third graph. Sometimes, when the parameter t does represent a quantity like time, we might indicate the direction of movement on the graph using an arrow, as shown above. There is often no single parametric representation for a curve. In Example 1 we assumed the object was moving at a steady rate along a straight line. If we kept the assumption about the path (straight line) but did not assume the speed was constant, we might get a system like: x(t ) = −5 + 2t 2
x(t)
y(t)
t
y (t ) = 3 − t 2 This starts at (-5, 3) when t = 0 and ends up at (3, -1) when t = 2. If we graph the x(t) and y(t) function separately, we can see that those are no longer linear, but if we graph x vs. y we will see that we still get a straight-line path.
y
t=0 t=1
x t=2
506 Chapter 8 Example 2 Sketch a graph of x(t ) = t 2 + 1 y (t ) = 2 + t We can begin by creating a table of values. From this table, we can plot the (x, y) points in the plane, sketch in a rough graph of the curve, and indicate the direction of motion with respect to time by using arrows. y
t -3 -2 -1 0 1 2
x 10 5 2 1 2 5
y -1 0 1 2 3 4
x
Notice that here the parametric equations describe a shape for which y is not a function of x. This is an example of why using parametric equations can be useful – since they can represent such a graph as a set of functions. This particular graph also appears to be a parabola where x is a function of y, which we will soon verify. While creating a t-x-y table, plotting points and connecting the dots with a smooth curve usually works to give us a rough idea of what the graph of a system of parametric equations looks like, it's generally easier to use technology to create these tables and (simultaneously) much nicer-looking graphs. Example 3 Sketch a graph of
x(t ) = 2 cos(t ) . y (t ) = 3 sin(t )
Using technology we can generate a graph of this equation, producing an ellipse. Similar to graphing polar equations, you must change the MODE on your calculator (or select parametric equations on your graphing technology) before graphing a system of parametric equations. You will know you have successfully entered parametric mode when the equation input has changed to ask for a x(t)= and y(t)= pair of equations.
y
x
Section 8.5 Parametric Equations 507 Try it Now
x(t ) = 4 cos(3t ) . This is an example of a Lissajous figure. y (t ) = 3 sin( 2t )
1. Sketch a graph of
20
100
15
80 Rabbits
Wolves
Example 4 The populations of rabbits and wolves on an island over time are given by the graphs below. Use these graphs to sketch a graph in the r-w plane showing the relationship between the number of rabbits and number of wolves.
10 5
60 40 20
0 0
1
2
3
4
5
6
0
7
0
Years
1
2
3
4
5
6
Years
For each input t, we can determine the number of rabbits, r, and the number of wolves, w, from the respective graphs, and then plot the corresponding point in the r-w plane. 100
Rabbits
80 60 40 20 0 0
5
10
15
20
Wolves
This graph helps reveal the cyclical interaction between the two populations. Converting from Parametric to Cartesian In some cases, it is possible to eliminate the parameter t, allowing you to write a pair of parametric equations as a Cartesian equation. It is easiest to do this if one of the x(t) or y(t) functions can easily be solved for t, allowing you to then substitute the remaining expression into the second part.
7
508 Chapter 8 Example 6 x(t ) = t 2 + 1 Write as a Cartesian equation, if possible. y (t ) = 2 + t Here, the equation for y is linear, so is relatively easy to solve for t. Since the resulting Cartesian equation will likely not be a function, and for convenience, we drop the function notation. Solve for t y = 2+t Substitute this for t in the x equation y−2=t x = ( y − 2) 2 + 1 Notice that this is the equation of a parabola with x as a function of y, with vertex at (1,2), opening to the right. Comparing this with the graph from Example 2, we see (unsurprisingly) that it yields the same graph in the x-y plane as did the original parametric equations. Try it Now 2. Write
x(t ) = t 3 y (t ) = t 6
as a Cartesian equation, if possible.
Example 7 x(t ) = t + 2 Write as a Cartesian equation, if possible. y (t ) = log(t ) We could solve either the first or second equation for t. Solving the first, x= t +2 Square both sides x−2= t 2 Substitute into the y equation (x − 2) = t
(
y = log ( x − 2 )
2
)
Since the parametric equation is only defined for t > 0 , this Cartesian equation is equivalent to the parametric equation on the corresponding domain. The parametric equations show that when t > 0, x > 2 and y > 0, so the domain of the Cartesian equation should be limited to x > 2. To ensure that the Cartesian equation is as equivalent as possible to the original parametric equation, we try to avoid using domain-restricted inverse functions, such as the inverse trig functions, when possible. For equations involving trig functions, we often try to find an identity to utilize to avoid the inverse functions.
Section 8.5 Parametric Equations 509 Example 8 x(t ) = 2 cos(t ) Write as a Cartesian equation, if possible. y (t ) = 3 sin(t ) To rewrite this, we can utilize the Pythagorean identity cos 2 (t ) + sin 2 (t ) = 1 . x x = 2 cos(t ) so = cos(t ) 2 y y = 3 sin(t ) so = sin(t ) 3 Starting with the Pythagorean Identity cos 2 (t ) + sin 2 (t ) = 1 2
Substitute in the expressions from the parametric form
2
x y + =1 2 3 x2 y2 + =1 4 9
Simplify
This is a Cartesian equation for the ellipse we graphed earlier. Parameterizing Curves While converting from parametric form to Cartesian can be useful, it is often more useful to parameterize a Cartesian equation – converting it into parametric form. If the Cartesian equation gives one variable as a function of the other, then parameterization is trivial – the independent variable in the function can simply be defined as t. Example 9 Parameterize the equation x = y 3 − 2 y . In this equation, x is expressed as a function of y. By defining y = t we can then substitute that into the Cartesian equation, yielding x = t 3 − 2t . Together, this produces the parametric form: x(t ) = t 3 − 2t y (t ) = t Try it Now 3. Write x 2 + y 2 = 3 in parametric form, if possible.
510 Chapter 8 In addition to parameterizing Cartesian equations, we also can parameterize behaviors and movements. Example 10 A robot follows the path shown. Create a table of values for the x(t) and y(t) functions, assuming the robot takes one second to make each movement. Since we know the direction of motion, we can introduce consecutive values for t along the path of the robot. Using these values with the x and y coordinates of the robot, we can create the tables. For example, we designate the starting point, at (1, 1), as the position at t = 0, the next point at (3, 1) as the position at t = 1, and so on. t x
0 1
1 3
2 3
3 2
4 4
5 1
6 1
t y
0 1
1 1
2 2
3 2
4 4
5 5
6 4
Notice how this also ties back to vectors. The journey of the robot as it moves through the Cartesian plane could also be displayed as vectors and total distance traveled and displacement could be calculated. Example 11 A light is placed on the edge of a bicycle tire as shown and the bicycle starts rolling down the street. Find a parametric equation for the position of the light after the wheel has rotated through an angle of θ.
θ
r Starting
Rotated by θ
Relative to the center of the wheel, the position of the light can be found as the coordinates of a point on a circle, but since the x coordinate begins at 0 and moves in the negative direction, while the y coordinate starts at the lowest value, the coordinates of the point will be given by: x = −r sin(θ ) y = − r cos(θ )
Section 8.5 Parametric Equations 511 The center of the wheel, meanwhile, is moving horizontally. It remains at a constant height of r, but the horizontal position will move a distance equivalent to the arclength of the circle drawn out by the angle, s = rθ . The position of the center of the circle is then x = rθ y=r Combining the position of the center of the wheel with the position of the light on the wheel relative to the center, we get the following parametric equationw, with θ as the parameter: x = rθ − r sin(θ ) = r (θ − sin(θ ) ) y = r − r cos(θ ) = r (1 − cos(θ ) ) The result graph is called a cycloid.
Example 12 A moon travels around a planet as shown, orbiting once every 10 days. The planet travels around a sun as shown, orbiting once every 100 days. Find a parametric equation for the position of the moon, relative to the center of the sun, after t days. For this example, we’ll assume the orbits are circular, though in real life they’re actually elliptical. The coordinates of a point on a circle can always be written in the form x = r cos(θ ) y = r sin(θ )
6 30
512 Chapter 8 Since the orbit of the moon around the planet has a period of 10 days, the equation for the position of the moon relative to the planet will be 2π π x(t ) = 6 cos t = 6 cos t 10 5 2π π y (t ) = 6 sin t = 6 sin t 10 5 With a period of 100 days, the equation for the position of the planet relative to the sun will be 2π π x(t ) = 30 cos t = 30 cos t 100 50 2π π y (t ) = 30 sin t = 30 sin t 100 50 Combining these together, we can find the position of the moon relative to the sun as the sum of the components. π π x(t ) = 6 cos t + 30 cos t 5 50 π π y (t ) = 6 sin t + 30 sin t 5 50 The resulting graph is shown here.
Try it Now 4. A wheel of radius 4 is rolled around the outside of a circle of radius 7. Find a parametric equation for the position of a point on the boundary of the smaller wheel. This shape is called an epicycloid. Important Topics of This Section Parametric equations Graphing x(t) , y(t) and the corresponding x-y graph Sketching graphs and building a table of values Converting parametric to Cartesian Converting Cartesian to parametric (parameterizing curves)
Section 8.5 Parametric Equations 513 Try it Now Answers
1. 2. y = x 2 3.
x(t ) = 3 cos(t ) y (t ) = 3 sin(t )
11 = x(t ) 11cos ( t ) − 4 cos t 4 4. 11 = y (t ) 11sin ( t ) − 4sin t 4
514 Chapter 8
Section 8.5 Exercises Match each set of equations with one of the graphs below. x ( t ) = 4sin ( t ) x ( t ) = t − 1 x ( t ) = t 1. 2. 3. 2 2 ) t −1 y ( t ) = t y ( t = y ( t ) = 2 cos ( t ) x ( t ) = 2sin(t ) 4. y ( t ) = 4 cos(t )
x ( t )= 2 + t 5. y ( t )= 3 − 2t
x ( t ) =−2 − 2t 6. y ( t )= 3 + t
A
B
C
D
E
F
From each pair of graphs in the t-x and t-y planes shown, sketch a graph in the x-y plane.
7.
8.
Section 8.5 Parametric Equations 515 From each graph in the x-y plane shown, sketch a graph of the parameter functions in the t-x and t-y planes.
9.
10.
Sketch the parametric equations for −2 ≤ t ≤ 2 . x ( t ) = 1 + 2t 11. 12. 2 y ( t ) = t
x ( t= ) 2t − 2 3 y ( t ) = t
Eliminate the parameter t to rewrite the parametric equation as a Cartesian equation x ( t )= 5 − t x ( t )= 6 − 3t 13. 14. ) 10 − t y ( t )= 8 − 2t y ( t= x ( t= ) 2t + 1 15. y ( t ) = 3 t
) 3t − 1 x ( t = 16. 2 y ( t ) = 2t
t x ( t ) = 2e 17. y ( t ) = 1 − 5t
x ( t ) = 4 log ( t ) 18. y ( t )= 3 + 2t
x ( t )= t 3 − t 19. y ( t ) = 2t
x ( t ) = t − t 4 20. y ( t ) = t + 2
x ( t ) = e 2t 21. 6t y ( t ) = e
x ( t ) = t 5 22. 10 y ( t ) = t
x ( t ) = 4 cos ( t ) 23. y ( t ) = 5sin ( t )
x ( t ) = 3sin ( t ) 24. y ( t ) = 6 cos ( t )
516 Chapter 8 Parameterize (write a parametric equation for) each Cartesian equation 25. y (= 26.= x ) 3x 2 + 3 y ( x ) 2sin ( x ) + 1 27. = x ( y ) 3log ( y ) + y 29.
x2 y 2 + = 1 4 9
28. x ( = y) 30.
y + 2y
x2 y 2 + = 1 16 36
Parameterize the graphs shown.
31.
32.
33.
34.
35. Parameterize the line from (−1,5) to (2,3) so that the line is at (−1,5) at t = 0, and at (2, 3) at t = 1.
36. Parameterize the line from (4,1) to (6, −2) so that the line is at (4,1) at t = 0, and at (6, 2) − at t = 1.
Section 8.5 Parametric Equations 517 x ( t ) = a cos ( bt ) The graphs below are created by parameteric equations of the form . y ( t ) = c sin ( dt ) Find the values of a, b, c, and d to achieve each graph.
37.
38.
39.
40.
41. An object is thrown in the air with vertical velocity 20 ft/s and horizontal velocity 15 ft/s. The object’s height can be described by the equation y ( t ) = −16t 2 + 20t , while the object moves horizontally with constant velocity 15 ft/s. Write parametric equations for the object’s position, then eliminate time to write height as a function of horizontal position. 42. A skateboarder riding on a level surface at a constant speed of 9 ft/s throws a ball in the air, the height of which can be described by the equation y ( t ) = −16t 2 + 10t + 5 . Write parametric equations for the ball’s position, then eliminate time to write height as a function of horizontal position.
518 Chapter 8
43. A carnival ride has a large rotating arm with diameter 40 feet centered 35 feet off the ground. At each end of the large arm are two smaller rotating arms with diameter 16 feet each. The larger arm rotates once every 5 seconds, while the smaller arms rotate once every 2 seconds. If you board the ride when the point P is closest to the ground, find parametric equations for your position over time. 44. A hypocycloid is a shape generated by tracking a fixed point on a small circle as it rolls around the inside of a larger circle. If the smaller circle has radius 1 and the large circle has radius 6, find parametric equations for the position of the point P as the smaller wheel rolls in the direction indicated.
P
P
519
Solutions to Selected Exercises Chapter 1 Section 1.1 1. a. f ( 40 ) = 13 b. 2 Tons of garbage per week is produced by a city with a population of 5,000. 3. a. In 1995 there are 30 ducks in the lake b. In 2000 there are 40 ducks in the late 5. a ,b, d, e 7. a, b 9. a, b, d 11. b 13. b, c, e, f 15. = f (1) 1, = f ( 3) 1 2 ) 4, g ( −= 3) 2 17. g (=
= f ( 3) 53, = f ( 2) 1 19.
f ( −2 ) 8 49 4 4 -4 5 1/4
35. a. -6
6 18 -1 4.414 -6 DNE 1/2
4 3 0 4.732 -6 -3 1
b.-16
39. a. iii b. viii c. I 41. a. iv b. ii c. v 2 43. ( x − 3) + ( y + 9) 2 = 36
d. ii d. I
height
age
b. a
f (1)
e. vi e. vi
f. iv f. iii
0 21 -4 5.236 0 -1/3 4
g. v
b. − h. vii
(b)
5 3
(c)
time
c. r
f ( 2)
2 4 1 5 -4 -1 2 37. a. 5
height of head
45. (a)
47a. t
f (0)
postage
21. 23. 25. 27. 29. 31. 33.
f ( −1)
d. L: (c, t) and K: (a, p)
weight
520
Section 1.2 1. D: [-5, 3)
R: [0,2]
3. D: 2 < t ≤ 8
R: 6 ≤ g ( t ) < 8
5. D: [0,4]
R: [-3, 0]
7. [2, ∞)
9. (−∞,3]
1 1 13. (−∞,− ) ∪ − , ∞ 2 2 17. (−∞,−11) ∪ (− 11,2 )∪ (2, ∞ )
11. (−∞,6) ∪ (6, ∞ ) 15. [ −4, 4 ) 4, ∪( ∞ ) f ( −1) 19. 21. 23.
-4 -1 -5
2 if 25. f ( x ) = − 2 if − 4 if
f (0) 6 -2 3
− 6 ≤ x ≤ −1 − 1 < x ≤ 2 2 < x ≤ 4
35.
20 7 3 if x ≤ 0 3 27. f ( x ) = 2 if x > 0 x
2 x + 3 if −3 ≤ x < −1 29. f ( x ) = x − 1 if −1 ≤ x ≤ 2 − 2 if 2 < x ≤ 5
31.
f ( 2)
33.
f ( 4) 34 5 16
521
Section 1.3 1. a) 6 million dollars per year b) 2 million dollars per year 3.
4−5 1 =− 4 −1 3
7. 27 11. 4b+4
5. 6 352 27 13. 3
9.
1 17. 9 + 9h + 3h 2 13h + 169 19. 4 x + 2h 21. Increasing: (− 1.5, 2 ) . Decreasing: (− ∞,−1.5) ∪ (2, ∞ )
15. −
23. Increasing: (− ∞,1) ∪ (3,4 ) . Decreasing: (1,3) ∪ (4, ∞ ) 25. Increasing, concave up
27. Decreasing, concave down
29. Decreasing, concave up
31. Increasing, concave down
33. Concave up (− ∞,1) . Concave down (1, ∞ ) . Inflection point at (1, 2) 35. Concave down (− ∞,3) ∪ (3, ∞ ) 37. Local minimum at (3, -22). Inflection points at (0,5) and (2, -11). Increasing on (3, ∞ ) . Decreasing (− ∞,3)
Concave up (− ∞,0 ) ∪ (2, ∞ ) . Concave down (0,2)
39. Local minimum at (-2, -2) Decreasing (− 3,−2 ) Increasing (− 2, ∞ )
Concave up (− 3, ∞ )
41. Local minimums at (-3.152, -47.626) and (2.041, -32.041) Local maximum at (-0.389, 5.979) Inflection points at (-2, -24) and (1, -15) Increasing (− 3.152,−0.389) ∪ (2.041, ∞ )
Decreasing (− ∞,−3.152) ∪ (− 0.389,2.041) Concave up (− ∞,−2 ) ∪ (1, ∞ ) Concave down (− 2,1)
522
Section 1.4 1. f ( g (0)) = 36 . g ( f (0)) = −57 3. f ( g (0)) = 4 . g ( f (0)) = 4 5. 4
7. 9
9. 4
21. f ( g ( x ) ) =
11. 7
13. 0
15. 4
x 7
17. 3
19. 2
g ( f ( x )= ) 7 x − 36
23. f ( g ( x )) = x + 3
g ( f (= x ))
25. f ( g ( x= )) 5x + 1
g ( f ( x= )) 5 x + 1
(
) (
27. f g ( h ( x ) ) =
)
x2 + 3
4
x −6 +6
29a. (0,2 ) ∪ (2, ∞ )
b. (− ∞,−2) ∪ (2, ∞ )
c. (0, ∞ )
31. b
33a. r (V ( t ) ) =
35. g ( x ) = x + 2, f ( x ) = x2
37. f ( x )=
3
3 (10 + 20t ) 4π
b. 208.94
3 g ( x )= x − , 5 x
39. f ( x ) = 3 + x , 2 g ( x) = x − , or f ( x ) = 3 + x, g ( x ) = x − 2 41a. f ( f ( x = b ) ) a ( ax + b ) += b. g ( x ) = 6 x −
8
( a ) x + ( ab + b ) 2
or g ( x ) = − 6 x −
6 +1
8 1− 6
2
s 70 60 43a. C ( f ( s ) ) = 2 s 10 + 60 c. v ( C ( m ) ) =
b. C ( g ( h ) ) =
70 ( 60h )
2
10 + ( 60h )
2
5280 70m 2 3600 10 + m 2
Section 1.5 1. Horizontal shift right 49 units
3. Horizontal shift left 3 units
523 5. Vertical shift up 5 units
7. Vertical shift down 2 units
9. Horizontal shift right 2 units, Vertical shift up 3 units 13. f ( x − 3) − 4 =
11. f ( x + 2 ) + 1 = x + 2 + 1
1 −4 x−3
15. g ( x ) = f ( x − 1) , h ( x ) =+ f ( x) 1
17.
19.
21.
23.
25. y = x − 3 − 2
27. y =
x + 3 −1
31. 33a. − f ( − x ) = −6− x
b. − f ( x + 2 ) − 3 =−6 x + 2 − 3
35. y = − ( x + 1) + 2 2
39a. Even
b. Neither
37. y = c. Odd
41. Reflect f(x) about the x-axis 43. Vertically stretch y values by 4
−x +1
29. y = − x
524 45. Horizontally compress x values by 1/5 47. Horizontally stretch x values by 3 49. Reflect f(x) about the y-axis and vertically stretch y values by 3 51. f ( −4 x ) = −4 x 53.
1 1 2) − 3 f ( x += −3 2 3 3( x + 2)
55. f (2( x − 5)) + 1 = (2( x − 5)) + 1 2
57. Horizontal shift left 1 unit, vertical stretch y values by 4, vertical shift down 5 units
becomes 59. Horizontal shift right 4 units, vertical stretch y values by 2, reflect over x axis, vertically shift up 3 units.
becomes 61. Vertically compress y values by ½
becomes
525 63. Horizontally stretch x values by 3, vertical shift down 3 units
becomes 65. Reflected over the y axis, horizontally shift right 4 units a ( x ) = − ( x − 4 )
becomes 67. This function is increasing on (−1, ∞) and decreasing on (−∞,−1) 69. This function is decreasing on (−∞,4) 71. This function is concave down on (−3, ∞) and concave up on (−∞,−3) 73. This function is concave up everywhere 75. f (− x )
77. 3 f ( x )
79. 2 f (− x )
1 81. 2 f x 2
83. 2 f ( x ) − 2
85. − f ( x + 1) + 3
87. y = −2 ( x + 2 ) + 3 2
93. y =
−1
( x − 2)
2
+3
99a. Domain : 3.5 ≤ x ≤ 6
3
1 89. y = ( x − 1) + 2 2
91. y = 2(x + 2 ) + 1
95. y =−2 x + 1 + 3
97. y = 3 −
d. Range : 9 − ≤ y≤7
1 (x − 2) + 1 2
526
Section 1.6 1. 6
3. -4
5. ½
7a. 3
b. 2
c. 2
d. 3
9a. 0
b. 7
c. 1
d. 3
11. x f
−1
(x )
1
4
7
12
16
3
6
9
13
14
13. f −1 ( x )= x − 3
15. f −1 ( x ) =− x + 2
17. f −1 ( x ) =
x−7 11
f −1 ( x= 19. Restricted domain x ≥ −7, 7 ) x− 21. Restricted domain x ≥ 0, f −1 ( x ) = x+5 23a. f ( g ( x )= )
(
3
)
3
x + 5 − 5= x
b. g ( f (x )) = 3 x 3 − 5 + 5 = x
c. This means that they are inverse functions (of each other)
Chapter 2 Section 2.1 1. = P ( t ) 1700t + 45000
3. D ( t= ) 10 + 2t
5. M (n ) = 40 − 2n
7. Increasing 9. Decreasing 11. Decreasing 13. Increasing 15. Decreasing 17. 3 1 4 2 21. 23. 19. − 3 3 5 25. - 0.05 mph (or 0.05 miles per hour toward her home) 27. Population is decreasing by 400 people per year 29. Monthly charge in dollars has an initial base charge of $24, and increases by $0.10 for each minute talked 31. Terry started at an elevation of 3,000 ft and is descending by 70ft per second. 1 11 3 33. y = x − 1 35. = 37. y = − x+ y 3x − 2 3 3 5
527 39. y = −1.5 x − 3
41. = y
45. P ( n ) = −0.004n + 34
2 x +1 3
47. The 1st ,3rd & 4th tables are linear: respectively 3. f ( x) = 5 x − 5 1. g ( x) = −3 x + 5 49a.= C
5 160 F− 9 9
b. = F
9 C + 32 5
43. y = −2 x + 3
4. k ( x) = 3 x − 2 c. − 9.4° F
Section 2.2 1. E
3. D
5. B
7.
9.
11.
13.
15.
17.
19.
21.
23. a. g ( x )= 25. y = 3
3 ( x + 2) − 4 4
b. ¾ c. -5/2
528 27. x = −3 Vertical Intercept 29. (0,2) 31. (0,-5) 33. (0,4) 35. Line 1: m = −10 37. Line 1: m = −2 2 39. Line 1: m = − 3
Horizontal Intercept (2,0) (5/3, 0) (-10,0) Line 2: m = −10 Line 2: m = 1 3 Line 2: m = 2 1 43. = y t +1 2
41. y = −5 x − 2 47. (1.2, 10)
Parallel Neither Perpendicular 45. (-1,1)
49. Plan B saves money if the miles are > 111
1 9
Section 2.3 1a. 696 people d. 305 people
b. 4 years e. P (= t ) 305 + 174t
c. 174 people per year f. 2219 people.
3a. C= ( x ) 0.15 x + 10 b. The flat monthly fee is $10 and there is an additional $0.15 fee for each additional minute used c. $113.05 5a. P= b. 6640 moose ( t ) 190t + 4170 7a. R ( t= ) 16 − 2.1t
b. 5.5 billion cubic feet
c. During the year 2017
9. More than 133 minutes
11. More than $42,857.14 worth of jewelry
13. 20.012 square units
15. 6 square units
b2 2m 19a. Hawaii 21. 26.225 miles 17. A = −
b. $80,640
c. During the year 1933
529
Section 2.4
1.
http://www.mathcracker.com/scatter_plot.php
3. y = 1.971x − 3.519, r = 0.967
5. y = −0.901x + 26.04, r = −0.968
7. 17.483 ≈ 17 situps
9. D
11. A
13. Yes, trend appears linear because r =0.994 and will exceed 35% near the end of the year 2019.
Section 2.5 1. y =
1 x + 2 +1 2
5.
3. y =−3 x − 3 + 3
7.
9 13 11. x = − or x = 5 5 5 1 15. x = − or x = − 3 3 Horizontal Intercepts 17. (-6, 0 ) and (4, 0) 19. none
9. 13. x =
21. −11 < x < 1 or (−11,1) 23. x ≥ 5 , x ≤ −1 or (−∞,−1] ∪ [5, ∞) 25. −
13 5 13 5 < x < − or (− ,− ) 3 3 3 3
1 15 = or x 2 2
Vertical Intercept (0, -8) (0, -7)
530
Chapter 3 Section 3.1 1. As x → ∞, f ( x) → ∞ As x → −∞, f ( x) → ∞ 3. As x → ∞, f ( x) → ∞ As x → −∞, f ( x ) → −∞ 5. As x → ∞, f ( x ) → −∞ As x → −∞, f ( x ) → −∞ 7. As x → ∞, f ( x ) → −∞ As x → −∞, f ( x) → ∞ th 9. 7 Degree, Leading coefficient 4 11. 2nd Degree, Leading coefficient -1 13. 4th Degree, Leading coefficient -2 15. 3rd Degree, Leading coefficient 6 17. As x → ∞, f ( x ) → −∞ As x → −∞, f ( x ) → −∞ As x → −∞, f ( x) → ∞ 19. As x → ∞, f ( x) → ∞ 21. intercepts: 5, turning points: 4 23. 3 25. 5 27. 3 29. 5 31. Horizontal Intercepts (1,0), (-2, 0), (3, 0) Vertical Intercept (0, 12) 33. Horizontal Intercepts (1/3, 0) (-1/2, 0) Vertical Intercept (0, 2)
Section 3.2 1. f ( x ) = ( x − 2 ) − 3
3. f (x ) = −2(x − 2 ) + 7
2
7. 9. 11.
2
5. f ( x ) =
Vertex ( −2.5, −0.5)
Vertical Intercept (0,12)
( 2.5, −8.5) ( 0.75,1.25)
(0,4)
(0.438, 0)
(4.562,0)
(0,-1)
(0.191, 0)
(1.309, 0)
13. f ( x ) =( x − 6 ) − 4 2
Horizontal Intercepts (-2, 0) (-3, 0)
15. f ( x ) = 2( x + 2 ) − 18
2 (x + 3)(x − 1) 3 1 2 23. f ( x ) = − ( x − 4) 4 27a. 234m b. 2909.561 ft 29a. 3 ft b. 111 ft 31. 24.91 in by 24.91 in 1 33. 125 ft by 83 ft 3 35. 24.6344 cm 37. $10.70
19. f ( x ) = −
1 2 ( x − 3) − 1 2
2
17. b = 32 and c = -39
3 ( x − 2 )( x − 5) 5 1 2 25. f ( x ) = − ( x + 3) + 2 9 c. 47.735 seconds c. 72.497 ft
21. f ( x ) =
531
Section 3.3 C(t) 1. 3. 5.
C, intercepts (0,48) (0,0) (0,0)
t, intercepts (4,0), (-1,0), (6,0) (0,0), (2,0), (-1,0) (0,0), (1,0), (3,0)
7. (-1.646, 0) (3.646, 0) (5,0) 9. As t → ∞, h(t ) → ∞ t → −∞, h(t ) → −∞ 11. As t → ∞, p(t ) → −∞ t → −∞, p(t ) → −∞
13.
15.
17. 19. (3, ∞) 23. [3.5, 6]
21. (− ∞, −2) ∪ (1,3) 25. (− ∞, 1]∪ (4, ∞ )
27. [−2,−2] ∪ [3, ∞ ) 2 31. y = − ( x + 2 )( x − 1)( x − 3) 3
35. y = −15 ( x − 1) ( x − 3) 2
3
39. y = −( x + 1) 2 ( x − 2) 1 (x + 4)(x + 2)(x − 3)2 24 1 3 47. y = ( x + 3)( x + 2 )( x − 1) 6 51. Base 2.58, Height 3.336
43. y =
29. ( −∞, −4 ) ∪ ( −4, 2 ) ∪(2, ∞)
1 2 33. y = ( x − 1) 2 ( x − 3) ( x + 3) 3 1 37. y = ( x + 2 )( x − 1)( x − 3) 2 1 41. y = − ( x + 3)( x + 2 )( x − 2 )( x − 4 ) 24 1 2 2 45. y = ( x + 2 ) ( x − 3) 12 1 2 49. y = − ( x + 3)( x + 1)( x − 2 ) ( x − 4 ) 16
532
Section 3.4 1. D
3. A
13.
Vertical Asymptotes x = −4 x=2 1 x = −4, 1 3 x = −1 , hole at x =1 x=4
15. 17.
x = 0, 4 x = −2, 4
5. 7. 9. 11.
Horizontal Asymptote y=2 y=0 y =1
Vertical yIntercept (0,-3/4) (0,1) (0, 5/16)
Horizontal xintercept (3/2, 0) DNE (-1/3, 0), (5,0)
y =1
(0,3)
(-3, 0)
none y=2x (oblique) y=0 y =1
(0, ¼)
(-1, 0), (1/2, 0)
DNE (0, -15/16)
(-2, 0), (2/3, 0) (1, 0), (-3, 0), (5, 0)
5.
7.
9.
11.
13.
15.
533
17. 50 ( x − 2 )( x + 1) 19. y = ( x + 5)( x − 5)
23. y =
1( x − 2 )
35.
2
27 ( x − 2 )
( x + 3)( x − 3) 2 − 6( x − 1) y= (x + 3)(x − 2)2 3 2 ( x − 1) y= 2 ( x + 1)( x − 2 )
2
39. a. C (n) =
7 ( x − 4 )( x + 6 )
( x + 4 )( x + 5) 4 ( x − 3) 25. y = ( x + 3)( x − 4 ) 1( x + 3)( x − 2 ) 29. y = 3 ( x − 1) 2 ( x )( x − 3) 33. y = − ( x + 3)( x − 4 ) (x − 4)(x − 2) 37. y = (x − 4)(x + 1)
2 ( x + 1)
27. y = 31.
21. y =
4 b. C (10) ≈ 13.33% c. 80 mL d. as n → ∞, C → 0 20 + n
Section 3.5 1. Domain ( 4, ∞ )
Inverse f −1 ( = x)
x +4
3. Domain ( −∞, 0 )
Inverse f −1 ( x ) = − 12 − x
5. Domain ( −∞, ∞ )
Inverse f −1 ( x ) =
7. = f ( x) −1
( x − 9)
2
4
2 − 8x x 5x − 4 15. f −1 (x ) = 3 + 4x 19. 34.073 mph
11. f −1 ( x ) =
+1
3
x −1 3 x −9 9. f ( x ) = 2 3 − 7x 13. f −1 ( x ) = x −1 −1
17. 65.574 mph 21. 14.142 feet
3
534
Chapter 4 Section 4.1 1. Linear t 7. P ( t ) = 11, 000 (1.085 )
3. Exponential 9. 47622 Fox
5. Neither
11. $17561.70
13. y = 6 ( 5 )
15. y = 2000 ( 0.1)
17. y = 3 ( 2 )
1 x x 1 5 1 5 19. y = = 2.93 ( 0.699 ) 21. y = ( 2 ) 8 6 6 25. 1.39%; $155,368.09 27. $4,813.55 Quarterly ≈ $7, 469.63 Monthly ≈ $7, 496.71
x
−
x
3
x
x
23. 34.32 mg 29. Annual ≈ $7353.84 Continuously ≈ $7,510.44 33. 7.4 years 31. 3.03% t 35a. w ( t ) = (1.113)(1.046 ) b. $1.11 c. Below what the model predicts ≈ $5.70
Section 4.2 1. B
3. A
5. E
11.
13.
17. = y 4x + 4
19. y = 4 x + 2
7. D
9. C
15. 21. y = −4 x
23. As x → ∞ f ( x ) → −∞ . As x → −∞ f ( x ) → −1
x → ∞ f ( x ) → −2 As x → −∞ f ( x ) → ∞ 25. As 27. As x → ∞ f ( x ) → 2
As x → −∞ f ( x ) → ∞
29. y = −2 x + 2 + 1 = −4(2) x + 1
31. y = −2(2) − x + 3
33. y = −2(3) + 7
1 35. y = 2 − 4 2
x
x
Section 4.3 1. 4m = q 7. e n = w
3. a c = b 9. log 4 ( y ) = x
13. log(b) = a
15. ln ( h ) = k
19. 1/8 25. 2
21. 1000 27. -3
5. 10t = v 11. log c (k ) = d 17. 9 23. e 2 29. ½
535 31. 4
33. -3
37. -1.398
39. 2.708
1 log 15 ≈ −1.392 43. log ( 7 )
49.
log ( 5 )
log (1.03)
45.
≈ 54.449
5 log 8 ≈ 0.678 55. 1 log 2 61. f ( t ) = 150 (1.0618 )
t
67. During the year 2074
ln (17 ) 5
35. -2 log(14 ) 41. ≈ 1.6397 log(5) log(38) +5 log(3) 47. ≈ 2.078 4
≈ 0.567
8 log 3 ≈ 8.335 51. 3 log(1.04 )
1 ln 5 53. ≈ 13.412 −0.12
57. f ( t ) = 300e −0.0943t
59. f ( t ) = 10e0.03922t
63. f ( t ) = 50 ( 0.98807 )
t
69. ≈ 34 hours
71. 13.532 years
Section 4.4 1. log 3 ( 4 )
3. log 3 (7 )
5. log 3 (5)
65. During the year 2013
7. log 7 (2 )
( )
9. log 6 x 9
xz 3 13. log x ( x + 1) 15. log 11. ln 2x y 17. 15log ( x ) + 13log ( y ) − 19 log ( z ) 19. −2 ln ( a ) + 4 ln ( b ) − 5ln ( c )
(
( ) 7
3 log ( x ) − 2log( y ) 2 8 14 25. log( x ) + log( y ) 3 3 27. x ≈ −0.717 2 33. x = 7 39. x ≈ 30.158 12 45. = x ≈ 1.091 11
)
23. ln ( y ) +
21.
1 ( ln ( y ) − ln (1 − y ) ) 2
29. x ≈ −6.395
31. t ≈ 17.329
35. x ≈ 0.123
37. x ≈ 4.642
41. x ≈ −2.889 .
43. x ≈ 6.873 or x ≈ −0.873
47. x = 10
Section 4.5 1. Domain: : x > 5 3. Domain: x < 3 1 5. Domain: x > − 3
3
2
V. A. @ x = 5 V.A. @ x = 3 V.A. @ x = −
1 3
536 7. Domain: x < 0
V.A. @ x = 0
9.
11.
13.
15.
1 log(− ( x − 1)) log(2 ) 3 21. y = log ( x + 2 ) log ( 4 ) 17. y =
3 − log ( x + 4 ) 19. y = log ( 3) 2 23. y = − log ( − ( x − 5 ) ) log ( 5 )
Section 4.6 1. f ( t ) = 13 ( 0.9195 ) . t
2 mg will remain after 22.3098 minutes
3. f ( t ) = 200 ( 0.999564 ) . t
f (1000 ) = 129.3311 mg
5. r = -0.06448. Initial mass: 9.9018 mg. After 3 days: 0.01648 mg 7. f ( t ) = 250 ( 0.9909 ) . Half-life = 75.8653 minutes t
9. f ( t ) = a ( 0.999879 ) . 60% (0.60a) would remain after 4222.813 years t
11. P ( t ) = 1500 (1.02337 ) (t in minutes). After 2 hours = 24000. After 100 minutes = t
15119 13. a) 610.5143 (about 611) b) 25.6427 minutes
c) 10431.21
d) 106.9642 minutes
15. 23.1914 years 17. 53.319 hours 19. T ( t ) 90 ( 0.99166 ) + 75 . a) 134.212 deg b) 112.743 minutes = t
537
21. a)
b) 100 c) 269.487 d) 7.324 years
23. log ( x ) = −0.5 . x = 0.3162
25. log ( x ) = 1.5 . x = 31.623
Whisper
27.
10-10 10-9
Vacuum 10-8 10-7 10-6 10-5 10-4
29. 63095.7 times more intense
Jet 10-3 10-2 10-1 100
101
102
31. MMS magnitude 5.817
33. a) about 1640671 b) 1.4 hours c) No, because (2.042727 )0.693147 ≈ e 0.495105 d) Anja’s data predicts a continuous growth rate of 0.4116, which is much smaller than the rate 0.495105 you calculated. Our model would overestimate the number of cells. 35. a) The curve that increases rapidly at first is M(p) b) H(100) = 0.9775 c) Myoglobin: M(20) = 0.9524. Hemoglobin: H(20) = 0.3242 d) At 20 torrs: 0.6282. At 40 torrs: 0.2060. At 60 torrs: 0.0714
Efficiency seems to be maximized at about 8 torr 37. a) C ( t ) = 1.03526t , or C ( t ) = e0.03466t b) Volume of one cell:
4 π 50 ×10−4 3
(
)
3
≈ 5.236 ×10−7 cm3, so will need about
C ( t ) 1.9099 ×106 after 417.3 hours 1.9099 ×106 cells for a volume of 1cm3. = 39. 31.699 days
538
Section 4.7
1. log( f ( x) ) = log(1.3)x + log(4 )
3. log( f ( x) ) = log(0.2 )x + 1
. 1
5. y = e 2
x −1
1
x
= e −1e 2 ≈ 0.368(1.6487) x
7. y = 10 − x − 2 = 10 −210 −1x = 0.01(0.1) x 9. y = 776.682 (1.426 )
x
11. y = 731.92(0.738) x
13. Expenditures are approximately $205 15. y 7.599 = = (1.016 ) r 0.83064 , x
y =0.1493 x + 7.4893, r =0.81713 . Using the
better function, we predict electricity will be 11.157 cents per kwh
539
Chapter 5 Section 5.1 3. ( x − 8 ) + ( y + 10 ) = 82 2
1. 10
2
5. ( x − 7 ) + ( y + 2 ) = 293
7. ( x − 5 ) + 1 3 ( y − 8) =
9. 11. (0, 3 + 5) and (0, 3 − 5) 15. (-1.07335, 2.8533)
13. (1.3416407865, 7.683281573) 17. 29.87 miles
2
2
2
Section 5.2 70° 30°
-135° 1. 3. π
300°
5. 150° 7. 325° 8π π 9. 54° 11. 13. 9 2 15. 35 miles 17. 8π cm 19. 5.7596 miles 21. 28.6479° 23. 14.1372 cm2 25. 3960 rad/min 630.254 RPM 27. 2.094 in/sec, π/12 rad/sec, 2.5 RPM 29. 75,398.22 mm/min = 1.257 m/sec 31. Angular speed: π/12 rad/hr. Linear speed: 1036.73 miles/hr
2
540
Section 5.3 1. a. III b. II
3. −
4 3 7
7. −
5. − 9.
4 5
55 8
a. reference: 45°. Quadrant III. sin ( 225° ) =−
2 2 . cos ( 225° ) =− 2 2
b. reference: 60°. Quadrant IV. sin ( 300° ) =−
3 1 . cos ( 300° ) = 2 2
c. reference: 45°. Quadrant II. sin (135° ) =
2 2 . cos (135° ) =− 2 2
1 3 d. reference: 30°. Quadrant III. sin ( 210° ) =− . cos ( 210° ) =− 2 2 11.
a. reference: b. reference:
5π . Quadrant III. sin 4 4
2 5π . cos = − 2 4
π
7π . Quadrant III. sin 6 6
3 1 7π = − = − . cos 2 2 6
5π . Quadrant IV. sin 3 3
3 5π . cos = − 2 3
π
c. reference: d. reference: 13.
π
3π . Quadrant II. sin 4 4
π
3π a. sin − 4
2 − = 2
23π b. sin 6
1 = − 2
π c. sin − = −1 2
d. sin ( 5π ) = 0 2π b. 100° 3 5π 17. a. b. 280° 3 19. (-11.491, -9.642)
15. a.
3π cos − 4 23π cos 6
2 − = 2
3 = 2
cos ( 5π ) = −1
c. 220°
5π 3 2π d. 3
d.
1 = 2
2 2 3π . cos = − = 2 2 4
π cos − = 0 2
c. 40°
2 = − 2
e. 235° e. 55°
541
Section 5.4 1. sec (θ ) = 2 , csc (θ ) = 2 , tan (θ ) = 1 , cot (θ ) = 1 3. sec (θ ) = −
2 3 3 , csc (θ ) = 2 , tan (θ ) = − , cot (θ ) = − 3 3 3
5. sec (θ ) = −2 , csc (θ ) =
2 3 3 , tan (θ ) = − 3 , cot (θ ) = − 3 3
7. a. sec (135° ) =− 2 b. csc ( 210° ) =−2 c. tan ( 60° ) = 3 .
d. cot ( 225° ) =1
9. cos (θ ) = −
7 4 7 3 7 7 4 , sec (θ ) = − , csc (θ ) = , tan (θ ) = − , cot (θ ) = − 7 4 7 3 3
11. sin (θ ) = −
2 2 2 3 2 , csc(θ ) = − , sec(θ ) = −3 , tan (θ ) = 2 2 , cot (θ ) = 4 3 4
13. sin (θ ) =
5 13 12 13 5 , cos (θ ) = , sec (θ ) = , csc (θ ) = , cot (θ ) = 13 12 5 13 12
15. a. sin(0.15) = 0.1494 b. sin(4) = -0.7568 c. sin(70°) = 0.9397 d. sin(283°) = -0.9744
cos(0.15) = 0.9888 cos(4) = -0.6536 cos(70°) = 0.3420 cos(283°) = 0.2250
17. sec(t )
21. tan(t )
19. tan(t )
tan(0.15) = 0.1511 tan(4) = 1.1578 tan(70°) = 2.7475 tan(283°) = -4.3315
23. cot(t )
25. ( sec(t ) )
2
Section 5.5
1.
sin ( A) =
sec( A) =
5 41 4 41 5 , cos( A) = , tan ( A) = 41 41 4
41 41 4 , csc( A) = , cot ( A) = 5 4 5
3. c= 14, b= 7 3, B= 60°
5. a= 5.3171, c= 11.3257, A= 28°
7. a= 9.0631, b= 4.2262, B= 25° 11. 836.2698 ft 15. 660.35 feet 19. 143.0427
9. 32.4987 ft 13. 460.4069 ft 17. 28.025 ft 21. 86.6685
542
Chapter 6 Section 6.1
1.
2.
3.
4.
5. Amp: 3. Period= 2. Midline: y= -4. = f ( t ) 3sin (π t ) − 4 π 6. Amp: 2. Period= 4. Midline: y= = -3. f ( t ) 2sin t − 3 2 1 7. Amp: 2. Period= 4π. Midline: y== 1. f ( t ) 2 cos t + 1 2 8. Amp: 3. Period= π. Midline: y= -1. = f ( t ) 3cos ( 2t ) − 1 2π 9. Amp: 2. Period= 5. Midline: y= 3. f ( t ) = −2 cos t+3 5 2π 10. Amp: 1. Period= 3. Midline: y= -1. f ( t ) = − sin t −1 3
π
, Shift: 4 left, Midline: y = 5 4 12. Amp: 4, Period = 4, Shift: 3 right, Midline: y = 7 2π 13. Amp: 2, Period = , Shift: 7 right, Midline: y = 4 3
11. Amp: 3, Period =
543 2π , Shift: 4 left, Midline: y = -2 5 15. Amp: 1, Period = 12, Shift: 6 left, Midline: y = -3 12 , Shift: 3 left, Midline: y = 6 16. Amp: 8, Period = 7 π 17. f ( x ) 4sin ( x + 1) = 5 π 18. f ( x ) 3sin ( x + 1) = 3 π 19. f ( x ) cos ( x + 2 ) = 5 π 20. f ( x ) = −2 cos ( x − 1) 3 π 21. D(t= ) 50 − 7 sin t 12 π 22. D(t= ) 68 − 12sin t 12 23. a. Amp: 12.5. Midline: y = 13.5. Period: 10 π b. h ( t ) = −12.5cos t + 13.5 5 c. h ( 5 ) = 26 meters 24. a. Amp: 17.5. Midline: y = 20.5. Period: 8 π b. h(t ) = −17.5 cos t + 20.5 4 c. h ( 4 ) = 38 meters
14. Amp: 5, Period =
Section 6.2 1. II 3. I 5. Period:
π
. Horizontal shift: 8 right 4 7. Period: 8. Horizontal shift: 1 left 9. Period: 6. Horizontal shift: 3 left
544
11.
13.
15. π 17. f ( x ) 2sec = 2
π 19. f ( x ) 2 csc = 4
x − 1
21. tan ( − x ) = 1.5
23. sec ( − x ) = 2
25. csc ( − x ) = 5
27. − csc ( x )
x +1
Section 6.3 1.
π
3. −
4 3π 7. 4 13. 1.9823
19. 27.
9.
π
π 6
4 15. -0.9273
π
21. −
4
25 − x 5
5.
2
29.
π 6
3x 9x 2 + 1
π 3
11. −
π
3 17. 44.427°
23.
2 10 7
25.
1 17
545
Section 6.4 1.
5π 7π , 4 4
π 5π 3
,
3
7π + 2π k , where k is an integer 4 4 7π 11π 11. + 2π k , + 2π k , where k is an integer 6 6 π 2π 5π 2π 13. + k, + k , where k is an integer 18 3 18 3 5π 2π 7π 2π 15. k, k , where k is an integer + + 12 3 12 3 π 5π 17. + π k , + π k , where k is an integer 6 6 π 2π 5π 2π 19. + k, k , where k is an integer + 4 3 12 3 21. 4 + 8k , where k is an integer 1 5 23. + 2k , + 2k , where k is an integer 6 6
9.
π
3.
5.
π
7.
2
π 3π 2
+ 2π k ,
25. 0.2734, 2.8682 31. 0.7813, 5.5019 37. 0.9291, 3.0709
27. 3.7603, 5.6645 33. 0.04829, 0.47531 39. 1.3077, 4.6923
29. 2.1532, 4.1300 35. 0.7381, 1.3563
Section 6.5 1. c = 89 , A = 57.9946°, B = 32.0054° 3. b = 1 76 , A = 27.8181°, B = 62.1819° π 5. = y ( x ) 6sin ( x − 1) + 4 2 π 7. D ( t ) = 50 − 13cos ( t − 5 ) 12 π π 9. a. P (= t ) 129 − 25cos t b. P ( t ) = 129 − 25cos (t − 3) 6 6 11. 75 degrees 13. 8 15. 2.80869431742 17. 5.035 months
,
2
546
Chapter 7 Section 7.1 1.
7π 11π , 6 6
5.
2 10 + 8k , and + 8k , where k is an integer 3 3
7.
5π 7π + kπ and + kπ , where k is an integer 12 12
3.
π 5π 3
,
3
9. 0.1339 + 10k and 8.6614 + 10k , where k is an integer 11. 1.1438 + 13.
π 3π ,
2
2
17. 0, , π
2π 2π k and 1.9978 + k , where k is an integer 3 3
π 5π 3
15. 0.056, 1.515, 3.197, 4.647
, 0.644, 2.498 ,
3
19.
21. 1.183, 1.958, 4.325, 5.100 25. π ,
23.
π 5π 3
,
37. 41.
π 2π 3
,
3
,π ,
,
3
,
,
6
,
6
,
6
3π 7π 11π , , 2 6 6
6
,
3
35. 0,
π 3π 4
,
4
π , ,
5π 7π , 4 4
39. 0, π ,1 .231, 5.052
π 5π 3
,
31. 3.305, 6.120
4π 5π , 3 3
π 2π 5π 4π 6
6
27. 1.823, 4.460
3
29. 2.301, 3.983, 0.723, 5.560 33. 0,
π 5π 7π 11π
3
Section 7.2 1.
2+ 6 4
3.
− 2− 6 4
5.
2− 6 4
7.
2+ 6 4
547
9.
3 1 sin ( x ) − cos ( x ) 2 2
11. –
3 1 cos ( x ) + sin ( x ) 2 2
13. sec ( t )
15. tan ( x )
17. 8 ( cos ( 5 x ) − cos ( 27 x ) )
19. sin ( 8 x ) + sin ( 2 x )
21. 2 cos ( 5t ) cos ( t )
23. 2sin ( 5 x ) cos ( 2 x )
5 15 −2 − 5 3 2 1 25. a. − + − = 12 3 4 3 4
5 1 2 15 5 + 2 15 b. − 3 − 4 + 3 4 = 12 27. 0.373 +
2π 2π k and 0.674 + k , where k is an integer 3 3
29. 2π k , where k is an integer 31.
π
33.
7
+
π 4π 4π 3π 4π 4π k , and π + k, k, + k , where k is an integer + 3 3 7 7 7 3
7π 11π π +πk , + π k , and k , where k is an integer 12 12 4
35. 2 13sin( x + 5.3004) 37.
or 2 13sin( x − 0.9828)
29sin(3 x + 0.3805 )
39. 0.3681, 3.8544
41. 0.7854, 1.8158
43. tan ( 6t )
Section 7.3 1. a.
3 7 32
b.
3 7 31 c. 31 32
3. cos ( 56° )
5. cos(34°)
7. cos (18x )
9. 2sin (16x )
11. 0, π , 2.4189, 3.8643
13. 0.7297, 2.4119, 3.8713, 5.5535
548 15.
π π 5π 3π , , , 6 2 6 2
17. a.
19.
2π 4π 2π 4π 8π 10π 14π 16π ,0, , , , , , , 3 3 9 9 9 9 9 9
1 + cos (10 x ) 2
21.
3 1 1 − cos(16 x ) + cos(32 x ) 8 2 8
23.
1 1 1 1 − cos ( 2 x ) + cos ( 4 x ) − cos ( 2 x ) cos ( 4 x ) 16 16 16 16
25. a.
1 2 3 + 2 7
b.
1 2 3 − 2 7
c.
1 7−4 3
Section 7.4 π 1. y 3sin ( x − 3) − 1 = 6
3. Amplitude: 8, Period:
1 second, Frequency: 3 Hz (cycles per second) 3
π 40 5. P ( t ) = −19 cos t + t + 650 6 3
t π 7. P ( t ) = −33cos t + 900 (1.07 ) 6
9. D ( t ) = 10 ( 0.85 ) cos(36π t )
11. D ( t ) = 17 ( 0.9145 ) cos ( 28π t )
13. a. IV b. III
x π 15. = y 6 ( 4 ) + 5sin 2
π 17. y = −3sin + 2 x + 7 2
π 1 19. y = 8 cos x + 3 2 2
t
t
x
x
549
Chapter 8 Section 8.1 60°
10
12.26
25° 6
120
50°
70°
11.305
1.
9.059
3.
35° 4.421
23.222° 5
65.952°
65° 6.046
5.
25
6
49.048° 7.
9. β = 68°, a = 14.711, c = 20.138
1
40° 116.668° 11.042
11. β = 28.096°, γ = 32.904°, c = 16.149
13. Not possible. 15. β = 64.243°, γ = 72.657°, c = 257.328 OR β =115.757°, γ = 21.143°, c = 97.238
20
17.
60°
30.510° 20
28
76.102° 43.898° 24.980
13 19.
112.620° 11
21. c = 2.066,α = 52.545°, β = 86.255° 23. a = 11.269, β = 27.457°, γ = 32.543° 25. 177.562
27. 978.515 ft
29. Distance to A: 565.258 ft. Distance to shore: 531.169 ft 31. 529.014 m
33. 173.877 feet
35. 4.642 km, 2.794 km
37. 757.963 ft
39. 2371.129 miles
41. 65.375 cm2
43. 7.72
36.870°
550
Section 8.2 7 3 7 1. − ,− 2 2
3. 2 2, −2 2
5. 3 2, −3 2
7. (0,3)
(
(
)
)
3 3 3 9. − 2 , − 2
11. (−1.248, 2.728)
13. 2 5, 0.464
15. 2 13, 2.159
21. r = 3 sec(θ )
23. r =
25. r = 4sin (θ )
27. r =
29. x 2 + y 2 = 3y
31. y + 7 x = 4
33. x = 2 37. A
35. x 2 + y 2 =x + 2 43. C 45. D
( 17. (
) 34,5.253)
39. C
( 19. (
41. E
) 269, 4.057 ) sin (θ )
4 cos 2 (θ ) cos (θ )
( cos (θ ) − sin (θ ) ) 2
2
49.
51.
53.
55.
57.
59.
47. F
551
61.
63.
65.
Section 8.3 1. 3i
3. −12
5. 1 + 3i 9. −11 + 4i 13. 30 − 10i
7. 8 − i 11. −12 + 8i 15. 11 + 10i 3 19. + 2i 2 1 18 23. − − i 25 25 27. i
17. 20 3 5 + i 2 2 25. −1 29. 3cos ( 2 ) + 3sin ( 2 ) i = −1.248 + 2.728i
21.
3 2 3 2 – i 2 2
31. 3 3 + 3i
33. –
35. 6e0i
37. 4e 2
3π
39. 2 2e 34e
43.
π 4
i
0.540 i
47. 17e 4.467 i 5π
51. 6e 12
55. 1024e 59. 4096 63. 1.771 + 0.322i 5
3π i 4
45. 10e 2.820i 49.
26e6.086i
53. 2e 12 5π i 2
65.
41. 3 2e
7π
i
i
π
i
i
57. 4e 3 61. 0.788 + 1.903i
2 ≈ 1.149, 0.355 + 1.092i, −0.929 + 0.675i, −0.929 − 0.675i, 0.355 − 1.092i
1 3 1 3 1 3 1 3 67. 1, + i, − + i, 1, − − − i, − i 2 2 2 2 2 2 2 2
552
Section 8.4 1. −4, 2
3.
5. 3v − u
The vectors do not need to start at the same point
9. −6.128, −5.142
7. 3 2,3 2 11. Magnitude: 4, Direction: 90°
13. Magnitude: 7.810, Direction: 39.806° 15. Magnitude: 2.236, Direction: 153.435° 17. Magnitude: 5.385, Direction: 291.801° 19. Magnitude: 7.211, Direction: 236.310° 21. u + v = 3,2 , u − v = 1,−8 , 2u − 3v = 1,−21 23. 4.635 miles, 17.764 deg N of E 25. 17 miles. 10.318 miles 27. Fnet =−4, −11 29. Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North 31. 4.924 degrees. 659 km/hr 33. 4.424 degrees 35. (0.081, 8.602) 37. 21.801 degrees, relative to the car’s forward direction
553
Section 8.5 1. C
3. E
5. F
7.
9. x(t)
y(t)
11. 13. y =−2 + 2 x 17. x = 2e
1− y 5
x or y = 1 − 5ln 2
15. y = 3
3
y y 19. = x − 2 2 2
21. y = x
3
x ( t ) = t 25. t ) 3t 2 + 3 y (=
x −1 2
2
x y 23. + = 1 4 5
x ( t ) 3log ( t ) + t = 27. . y ( t ) = t
554 x ( t ) = 2 cos ( t ) 29. y ( t ) = 3sin ( t )
x ( t ) = t 3 31. y ( t ) = t + 2
x ( t ) = t − 1 33. 2 y ( t ) = −t
x(t ) = −1 + 3t 35. y (t ) = 5 − 2t
x ( t ) = 4 cos ( 3t ) 37. y ( t ) = 6sin ( t )
x ( t ) = 4 cos ( 2t ) 39. y ( t ) = 3sin ( 3t )
2
x x 41. y ( x ) = −16 + 20 15 15
2π = x ( t ) 20sin 5 t + 8sin (π t ) 43. 2π y (t ) = 35 − 20 cos t − 8cos (π t ) 5
555
Index Absolute Value Functions, 146 Graphing, 147 Solving, 148 Solving Inequalities, 149 Ambiguous Case, 454 Amplitude, 357, 361 Angle, 307 Coterminal Angles, 308 Degree, 307 Radian, 310 Reference Angles, 327 Standard Position, 307 Angular Velocity, 315 Annual Percentage Rate (APR), 222 Annual Percentage Yield (APY), 224 Arclength, 309 Arcsine, Arccosine and Arctangent, 380 Area of a Sector, 314 Average Rate of Change, 34 Change of Base, 246, 253 Circles, 298, 470 Area of a Sector, 314 Equation of a Circle, 298 Points on a Circle, 299, 322 Polar Coordinates, 470 Coefficients, 158 Cofunction Identities, 345 Common Log, 244 Complex Conjugate, 482 Complex Number, 480 Complex Plane, 481 Component Form, 494 Composition of Functions, 49 Formulas, 51 Tables and Graphs, 50 Compound Interest, 222 Concavity, 41 Continuous Growth, 225 Correlation Coefficient, 141, 142 Cosecant, 333 Cosecant Function Domain, 372 Range, 372 Cosine, 321, 343, 356
Cotangent, 333 Cotangent Function Domain, 373 Period, 373 Range, 373 Coterminal Angles, 308 Damped Harmonic Motion, 445 Decreasing, 38 Degree, 158, 307 Difference of Logs Property, 253 Domain, 21 Double Angle Identities, 431 Double Zero, 178 Doubling Time, 273 Euler’s Formula, 484 Even Functions, 71 Exponential Functions, 215 Finding Equations, 219 Fitting Exponential Functions to Data, 292 Graphs of Exponential Functions, 232 Solving Exponential Equations, 247 Transformations of Exponential Graphs, 235 Exponential Growth or Decay Function, 216 Exponential Property, 253 Extrapolation, 139 Extrema, 38, 182 Function, 1 Absolute Value Functions, 146 Composition of Functions, 49 Domain and Range, 21 Exponential Functions, 215 Formulas as Functions, 7 Function Notation, 3 Graphs as Functions, 5 Horizontal Line Test, 7 Inverse of a Function, 90 Linear Functions, 99, 101 Logarithmic Functions, 242 One-to-One Function, 2 Parametric Functions, 504 Periodic Functions, 353
556 Index Piecewise Function, 28 Polar Functions, 470 Power Functions, 155 Quadratic Functions, 163 Radical Functions, 206, 207 Rational Functions, 188, 191 Sinusoidal Functions, 355 Solving & Evaluating, 5 Tables as Functions, 3 Tangent Function, 369 Vertical Line Test, 6 Half-Angle Identities, 438 Half-Life, 270 Hole, 192 Horizontal Asymptote, 189, 194 Horizontal Intercept, 116 Horizontal Line Test, 7 Horizontal Lines, 117 Imaginary Number, 480 Complex Conjugate, 482 Complex Number, 480 Complex Plane, 481 Euler’s Formula, 484 Polar Form of Complex Numbers, 483, 484 Increasing, 38 Inflection Point, 41 Intercepts, 168, 176, 177, 181, 195 Graphical Behavior, 177 Writing Equations, 181 Interpolation, 139 Intersections of Lines, 119 Interval Notation, 23 Union, 23 Inverse of a Function, 90 Properties of Inverses, 94 Inverse Properties, 253 Inversely Proportional, 188 Inversely Proportional to the Square, 188 Inverses, 206 Law of Cosines General Pythagorean Theorem, 457 Law of Sines Ambiguous Case, 454 Leading Coefficient, 158 Leading Term, 158
Least-Square Regression, 140 Limaçons, 471 Linear Functions, 99, 101 Fitting Linear Models to Data, 138 Graphing, 111 Horizontal Intercept, 116 Horizontal Lines, 117 Intersections of Lines, 119 Least-Square Regression, 140 Modeling, 126 Parallel Lines, 117 Perpendicular Lines, 118 Veritcal Lines, 117 Vertical Intercept, 112 Linear Velocity, 315 Lissajous Figure, 507 Local Maximum, 38 Local Minimum, 38 Logarithmic Functions, 242 Change of Base, 246, 253 Common Log, 244 Difference of Logs Property, 253 Exponential Property, 245, 253 Graphs of Logarithmic Functions, 262 Inverse Properties, 242, 253 Logarithmic Scales, 276 Log-Log Graph, 290 Moment Magnitude Scale, 279 Natural Log, 244 Orders of Magnitude, 279 Semi-Log Graph, 290 Sum of Logs Property, 253 The Logarithm, 242 Transformations of the Logarithmic Function, 264 Log-Log Graph, 290 Long Run Behavior, 156, 159, 189 Mathematical Modeling, 99 Midline, 358, 361 Model Breakdown, 140 Moment Magnitude Scale, 279 Natural Log, 244 Negative Angle Identities, 410 Newton's Law of Cooling, 275 Nominal Rate, 222 Odd Functions, 71
557 One-to-One Function, 2, 7 Orders of Magnitude, 279 Parallel Lines, 117 Parametric Functions, 504 Converting from Parametric to Cartesian, 507 Lissajous Figure, 507 Parameterizing Curves, 509 Period, 353, 361 Periodic Functions, 353 Period, 353 Sinusoidal, 355 Perpendicular Lines, 118 Phase Shift, 364 Piecewise Function, 28 Polar Coordinates Converting Points, 468 Polar Functions, 470 Converting To and From Cartesian Coordinates, 473 Euler’s Formula, 484 Limaçons, 471 Polar Form of Complex Numbers, 483, 484 Roses, 471, 472 Polynomial, 158 Coefficients, 158 Degree, 158 Horizontal Intercept, 178, 181 Leading Coefficient, 158 Leading Term, 158 Solving Inequalities, 179 Term, 158 Power Functions, 155 Characterisitcs, 156 Power Reduction Identities, 438 Product to Sum Identities, 423 Pythagorean Identity, 323, 337 Alternative Forms, 337, 410 Pythagorean Theorem, 297 Quadratic Formula, 170 Quadratic Functions, 163 Quadratic Formula, 170 Standard Form, 165, 166 Transformation Form, 165 Vertex Form, 165
Radian, 310 Radical Functions, 206, 207 Range, 21 Rate of Change, 34 Average, 34 Using Function Notation, 36 Rational Functions, 188, 191 Hole, 192 Intercepts, 195 Long Run Behavior, 193 Reciprocal Identities, 410 Reference Angles, 327 Roses, 471, 472 Secant, 333 Secant Function Domain, 372 Range, 372 Semi-Log Graph, 290 Set-Builder Notation, 23 Short Run Behavior, 159, 163, 168, 176, 189 Sine, 321, 343, 356 Single Zero, 178 Sinusoidal Functions, 355 Amplitude, 357, 361 Damped Harmonic Motion, 445 Midline, 358, 361 Modeling, 399 Period, 353, 361 Phase Shift, 364 Solving Trig Equations, 393 Slope, 101, 102, 112 Decreasing, 101 Increasing, 101 Standard Form, 165, 166 Standard Position, 307 Sum and Difference Identities, 417 Sum of Logs Property, 253 Sum to Product Identities, 424 Tangent, 333, 343 Tangent Function, 369 Domain, 369 Period, 369 Range, 369 Term, 158 The Logarithm, 242
558 Index Toolkit Functions, 11 Domains and Ranges of Toolkit Functions, 26 Transformation Form, 165 Transformations of Functions, 61 Combining Horizontal Transformations, 77 Combining Vertical Transformations, 76 Horizontal Reflections, 68 Horizontal Shifts, 64 Horizontal Stretch or Compression, 74 Vertical Reflections, 68 Vertical Shifts, 62 Vertical Stretch or Compression, 72 Trigonometric Identities, 334 Alternative Forms of the Pythagorean Identity, 337, 410 Cofunction Identities, 345 Double Angle Identities, 431 Half-Angle Identities, 438 Negative Angle Identities, 410 Power Reduction Identities, 438 Product to Sum Identities, 423 Pythagorean Identity, 323 Reciprocal Identities, 410 Sum and Difference Identities, 417
Sum to Product Identities, 424 Trigonometry Cosecant, 333 Cosine, 321, 343, 356 Cotangent, 333 Right Triangles, 343 Secant, 333 Sine, 321, 343, 356 SohCahToa, 343 Solving Trig Equations, 393 Tangent, 333, 343 The Pythagorean Theorem, 297 Unit Circle, 328 Triple Zero, 178 Unit Circle, 328 Vector, 491, 494 Adding Vectors Geometrically, 492 Adding, Subtracting, or Scaling Vectors in Component Form, 496 Geometrically Scaling a Vector, 493 Vertex, 163, 165 Vertex Form, 165 Vertical Asymptote, 189, 192 Vertical Intercept, 112 Vertical Line Test, 6 Vertical Lines, 117
